The elements of geometrie of the most auncient philosopher Euclide of Megara. Faithfully (now first) translated into the Englishe toung, by H. Billingsley, citizen of London. Whereunto are annexed certaine scholies, annotations, and inuentions, of the best mathematiciens, both of time past, and in this our age. With a very fruitfull præface made by M. I. Dee, specifying the chiefe mathematicall scie[n]ces, what they are, and wherunto commodious: where, also, are disclosed certaine new secrets mathematicall and mechanicall, vntill these our daies, greatly missed

¶The sixth booke of Eu∣clides Elementes. (Book 6)

THIS SIXTH BOOKE, is for vse and practise,* 1.1 a most speciall booke. In it are taught the proporti∣ons of one figure to an other figure, & of their sides the one to the other, and of the sides of one to the sides of an other, likewise of the angles of the one to the angles of the other. Moreouer it teacheth the description of figures like to ••igures geuen, and mar∣ueilous applications of figures to lines, euenly, or with decrease or excesse, with many other Theo∣remes, not onely of the propo••tions of right lined fi∣gures, but also of sectors of circles, with their angles. On the Theoremes and Problemes of this Booke depend for the most part, the compositions of all instrumentes of measuring length, breadth, or de••pe••es, and also the reason of the vse of the same instrumentes, as of the Geometricall ••quar••,* 1.2 the Scale of the Astrolabe, the quadrant, the staffe, and such other. The vse of which instrumentes, besides all other mechanicall instrumentes of raysing vp, of mouing, and drawing huge thinges incredible to the ignorant, and infinite other ginnes (which likewise haue their groundes out of this Booke) are of wonderfull and vnspeakeable profite, besides the inestimable pleasure which is in them.

Definitions.

1. Like rectiline figures are such,* 1.3 whose angles are equall the one to the other, and whose sides about the equall angles are proportionall.

As if ye take any

[illustration]

two rectiline figures. As for example, two triangles ABC, and DEF: 〈…〉〈…〉 of the one triangle be e∣quall to the angles of the other, namely, if the angle A be equall to the angle D, and the angle B equall to the angle E, & also the an∣gle C equall to the an∣gle F. And moreouer, i•• the sides which con∣taine the equall angles be proportionall. As if the side AB haue that proportion to

the side BC, wh••ch the side DE hath to the side EF, and also if the side BC be vnto the side CA, as ••he side EF is to the side FD, and mor••ouer if the side CA be to the side AB, as the side FD is to the side DE, then are these two triangles sayd to be like: and so iudge ye of any other kinde of figures. As if in the paralle∣logrammes ABCD and EFGH, the angle A be equall to the angle E, and the angle B equall to the angle F, and the angle C equall to the angle G, and the an∣gle D equall to the angle H. And farthermore, if the side AC haue that propor∣tion to the side CD which the side EG hath to the side GH, and if also the side CD be to the side DB as the side GH is to the side HF, and moreouer, if the side DB be to the side BA as the side HF is to the side FE, and finally, if the side BA be to the side AC as the side FE is to the side EG, then are these parallelo∣grammes like.

* 1.42. Reciprocall figures are those, when the terme•• of proporti∣on are both antecedentes and consequentes in either figure.

As if ye haue two parallelogrammes

[illustration]

ABCD and EFGH. If the side AB to the side EF, an antecedent of the first figure to a consequent of the second figure, haue mutually the same proportion, which the side EG hath to the side AC an antece∣dent of the second figure to a consequent of the first figure: then are these two figures Reciprocal. They are called of some, fi∣gures of mutuall sides, and that vndoubted∣ly not amisse nor vnaptly.* 1.5 And to make thys definition more plaine, Campane and Pestitarius, and others•• thus put it: Reciprocall figures, are when the sides of other 〈◊〉〈◊〉 mutually proportionall, as in the example and declaration before geuen. Among the barbarous they are called Mutekesia, reseruing still the Ara∣bike worde.

* 1.63. A right line is sayd to be deuided by an extreme and meane proportion, when the whole is to the greater part, as the grea∣ter part is to the lesse.

As if the line AB, be so deuided in the point

[illustration]

C, that the whole line AB haue the same pro∣portion, to the greater part thereof, namely, to AC, which the same greater part AC hath to the lesse part therof, namely, to CB, then is the line AB deuided by an extreme and meane proportion. Commonly it is called a line deuided by proportion ha••ing a meane and two extremes. How to deuide a line in such sort was taught in the 11. Proposition of the second Booke, but not vnder this forme of proportion.

4. The alitude of a figure is a perpendicular line drawen from the toppe to the base.* 1.7

As the altitude or hight of the triangle ABC, is the line AD being drawen perpendicularly from the poynt A, being the toppe or highest part of the triangle to the base therof BC. So likewise in other figures as ye see in the examples here set. That

[illustration]

which here ••ee calleth the alti∣tude or height of a fi∣gure, in the first booke in the 35. Proposi∣tion and certaine other following, he taught to be contayned within two equidi∣stant lines: so that figures to haue one altitude and to be contayned within two e∣quidistant lines, is all one. So in all these examples, if from the highest point of the figure ye draw an equidistant line to the base therof, and then frō that poynt draw a perpendicular to the same base that perpendicular is the altitude of the figure.

5. A Proportion is said to be made of two proportions or more, when the quantities of the proportions multiplied the one into the other, produce an other quantitie.* 1.8

Of addition of proportions, hath bene somewhat sayd in the declaration of the 10. definition of the fift booke: which in substance is all one with that which is here taught by Euclide. By the name of quantities of proportions, he vnderstan∣deth the denominations of proportions. So that to adde two proportions toge∣ther, or more, and to make one of them all, is nothyng els, but to multiply their quantities together, that is to multiply euer the denominator of the one by the denominator of the other.* 1.9 Thys is true in all kindes of proportion, whether it be of equalitie, or of the greater inequalitie, when the greater quantitie is referred to the lesse: or of the lesse inequalitie, when the lesse quantitie is referred to the grea∣ter: or of them mixed together. If the proportions be like, to adde two together is to double the one, to adde 3. like is to triple the one, and so forth in like propor∣tions, as was sufficiently declared in the declaration of the 10. and 11. definitions of the fift Booke. Where it was shewed, that if there be 3. quantities in like pro∣portion, the proportion of the first to the thyrd, is the proportion of the first to the second doubled: and if there be foure quantities in like proportion; the proportion of the f••••st ••o the fourth shall be the proportion of the first to the second ••••ipled•• which thing how to do was there taught likewyse in proportions vnlike, the proportion of the first extreme to the last is made of all the meane proportions set

betwene them. Suppose three quantities A, B, C, so that let A haue to B sesqui∣altera proportion, namely, 6. to 4. And let B to C haue sesquitertia proportion, namely, 4. to 3.* 1.10 Now the proportion of A to

[illustration]

C, the first to the thyrd, is made of the propor∣tion of A to B, and of the proportion of B to to C added together. If ye will adde them to∣gether, ye must by this definition multiply the quantitie or denominator of the one, by the quātitie or denominator of the other. Ye must first therefore seeke the denominators of these proportions, by the rule before geuen in the declaration of the definitions of the fift Booke. As if ye deuide A by B, namely, 6. by 4, so shall ye haue in the quotient 1 ••/•• for the denominator of the proportion of A to B: likewise if ye deuide B by C, namely, 4. by 3. ye shall haue in the quotient 1 ••/•• for the denominator of the proportion of B to C, now multiply these two denominators 1 ••/•• and 1 ••/•• the one into the other, by the rule before taught, namely, by multiplying the numerator of the one into the nu∣merator of the other, and also the denominator of the one into the denominator of the other: the numerator of 1 ••/•• or of ••/•• which is all one, is 3, the denomi∣nator is 2: the numerator of 1 ••/•• which reduced are ••/•• is 4,

[illustration]

the denominator is 3: then multiply 3. by 4, numerator by nu∣merator, so haue ye 12. for a new numerator: likewise multiply 2. by 3. denominator by denominator, ye shall produce 6. for a new denominator: so haue you produced 12. and 6, betwene which there is dupla proportion. Which proportion is also be∣twene A and C, namely, 6. to 3, the first quantitie to the third. Wherfore the pro∣portion of A to C is sayd to be made of the proportion of A to B and of the pro∣portion of B to C, for that it is produced of the multiplication of the quantitie or denominator of the one,* 1.11 into the quantitie or denominator of the other. And so of all others be they neuer so many. As in these examples in numbers here set 2. 3. 15. 18.* 1.12 In this example the lesse numbers are compared to the greater, as in the former the greater were compared to the lesse: the denominator of the proporti∣on of 2. to 3. is 2/•• that is, subsesquialtera, the denomination betwene 3. and 15. is ••/•• or 1/•• which is all one, that is, subquintupla, betwene 15. and 18. the denomi∣nation of the proportion is 1/6 that is, subsesquiquinta, multiply all these denomi∣nations together: first the numerators: 2. into 1. produce 2, then 2. into 5. produce 10: which shall be a new numerator. Then the de∣nominators:

[illustration]

3. into 5. produce 15: and 15. into 6. produce 90: which shall be a new denominator. So haue you brought forth 10/9•• or 1/9 which is pro∣portion subnoncupla: which is also the proportion of 2. to 18. Wherefore the proportion of 2. to 18. that is, of the extremes, namely, subnoncupla, is made of the proportions of 2. to 3: of 3. to 15: and of 15. to 18: namely, of subsesquialtera, subquintupla, and subses∣quiquinta.

An other example, where the greater inequalitie and the lesse inequalitie are mixed together 6. 4. 2. 3. the denomination of the proportion of 6. to 4, is 1 ••/••,* 1.13 of 4. to 2, is ••/••, and of 2. to 3, is ••/••: now if ye multiply as you ought, all these denominations together, ye shall produce 12. to 6, namely, dupla proportion.

Forasmuch as so much hath hetherto bene spoken of addition of proportions

it shall not be vnnecessary somewhat also to say of substraction of them.* 1.14 Where it is to be noted, that as addition of them, is made by multiplicatiō of their denomi∣nations the one into the other: so is the substraction of the one from the other done, by diuision of the denomination of the one by the denomination of the o∣ther. As if ye will from sextupla proportion subtrahe dupla proportion, take the denominations of them both. The denomination of sextupla proportion, is 6, the denomination of dupla proportion, is 2. Now deuide 6. the denomination of the one by 2. the denomination of the other: the quotient shall be 3: which is the denomination of a new proportion, namely, tripla: so that when dupla pro∣portion is subtrahed from sextupla, there shall remayne tripla proportion. And thus may ye do in all others.

6. A Parallelogramme applied to a right line, is sayd to want in forme by a parallelogramme like to one geuen: whē the pa∣rallelogrāme applied wanteth to the filling of the whole line, by a parallelogramme like to one geuen:* 1.15 and then is it sayd to exceede, when it exceedeth the line by a parallelogramme like to that which was geuen.

As let E be a Parallelogrāme

[illustration]

geuen, and let AB be a right line, to whom is applied the pa∣rallelogramme ACDF. Now if it want of the filling of the line AB, by the parallelogrāme DFGB being like to the pa∣rallelogramme geuen E, then is the parallelogramme sayd to want in forme by a parallelo∣gramme like vnto a parallelogramme geuen.

Likewise if it exceede, as the parallelogramme ACGD applyed to the lin•• AB•• if it exceede it by the

[illustration]

parallelogramme FGBD being like to the parallelo∣gramme F which was ge∣uen, then is the parallelo∣gramme ABGD, sayd to exceede in forme by a pa∣rallelogramme like to a pa∣rallelogramme geuen.

This definition is added by Flussates as it seemeth, it is not in any cōmon Greke booke abroad, nor in any Commentary. It is for many Theoremes following very necessary.

The 1. Theoreme. The 1. Proposition. Triangles & parallelogrammes which are vnder one & the self same altitude: are in proportion as the base of the one is to the base of the other.

SVppose that there be two triangles ABC and ACD, & two paralleiogrammes EC and CF. Which let be set vn∣der one and the selfe same altitude, or perpendicular line drawen from the toppe A to the base BD. Then I say that as the base BC is to the base CD so is the triangle ABC to the triangle ACD: and the parallelogramme EC to the parallelogramme CF. For forasmuch as the parallelogrammes.* 1.16 E

[illustration]

C and CF are vnder one and the selfe same altitude, therfore the lines EA and AF make both one right line, and so also do the lines BC and CD: and therefore the lyne EF is a parallel vnto the line BD. Produce the right line DB on ••che side directly to the pointes H, L (by the 2. peticion of the first) And vnto the base BC (by the 2. of the first) put as many equall lines as you will, as for example, two, namely, BG, and GH and vnto the base CD on the other side put as many equall as you did to the other base, which let be DK, and KL. Then draw these right lines AG, AH, AK, and AL.

And forasmuch as the lines CB, BG, and GH are equall the one to the o∣ther,* 1.17 therfore the triangles also AHG, AGB and ABC, are (by the 38. of the first) equall the one to the other. Wherfore how multiplex the base HC is to the base BC, so multiplex also is the triangle AHC to the triangle ABC. And by the same reason also, how multiplex the base LC is to the base DC, so multi∣plex also is the triangle ALC to the triangle ADC. Wherfore if the base HC be equall vnto the base CL, then (by the 38. of the first) the triangle AHC is equall vnto the triangle ACL. And if the base HC exceede the base CL, then also the triangle AHC excedeth the triangle ACL, and if the base be lesse, the triāgle also shall be lesse. Now then there are foure magnitudes, namely, the two bas••s BC and CD, and the two triangles ABC, and ACD, and to the base BC, and to the triangle ABC, namely, to the first and the third, are taken eque∣mul••iplices,

namely, the base HC and the triangle AHC, and likewise to y^e base CD and to the triangle ADC, namely, to the second and the fourth, are taken certaine other equemultiplices, that is, the base CL, and the triangle ALC. And it hath bene proued that if the multiplex of the first magnitude, that is, the base HC, do exceede the multiplex of the second, that is, the base CL, the multiplex also of the third, that is, the triangle AHC excedeth the multiplex of y^e fourth•• that is, the triangle ALC, and if the said base HC be equall to the said ba•••• CL, the triangle also AHC is equall to the triangle ALC, and if it be lesse it i•• lesse. Wherfore by the sixt defini••ion of y^e fifth, as the first of the foresaid magni∣tudes is to the second, so is the third to the fourth. Wherfore as the base BC is to the base CD, so is the triangle ABC to the triangle ACD.

And because (by the 41. of the first) the parallelogramme EC is double to the triangle ABC,* 1.18 and (by the same) the parallelogramme FC is double to the triangle ACD, therfore the parallelogrammes EC and FC are equemultipli∣ces vnto the triangles ABC and ACD. But the partes of equemultiplices (by the 15. of the fifth) haue one and the same proportion with thei•• equemultiplices. Wherfore as the triangle ABC is to the triangle ACD, so is the parallelograme EC to the parallelogramme FC. And forasmuch as it hath bene demonstrated, that as the base BC is to the base CD, so is the triangle ABC, to the triangle ACD, and as the triangle ABC is to the triangle ACD so is the parallelo∣gramme EC to the parallelogramme FC. Wherefore (by the 11. of the fifth) as the base BC is to the base CD, so is the parallelogramme EC to the parallelo∣gramme FC. The parallelogrammes may also be demonstrated a part by them∣selues as the triangles are, if we describe vpon the bases BG, GH, and DK & KL parallelogrammes vnder the self same altitude that the parallelogramme•• geuen are. Wherfore triangles and parallelogrammes which are vnder one and the selfe same altitude, are in proportio••, as the base of the one is to y^e base of the other: which was required to be demonstrated.

Here Flussates addeth this Corollary.

If two right lines being geuen, the one of them be deuided how so euer:* 1.19 the rectangle fi∣gures contayned vnder the whole line vndeuided, and eche of the segmentes of the line deui∣ded, are in proportion the one to the other, as the segmentes are the one to the other. For i∣maginyng the figures BA and AD in the former description, to be rectangled, the rectangle figures contayned vnder the whole right lyne AC, and the segments of the right line BD, which is cu•• in the poynt C, namely, the parallelogrammes BA and AD, are in proportion the one to the other, as the segmētes BC and CD are.

The 2. Theoreme. The 2. Proposition. If to any one of the sides of a triangle be drawen a parallel

right line, it shall cut the sides of the same triangle proportio∣nally. And if the sides of a triangle be cut proportionally, a right lyne drawn from section to section is a parallel to the o∣ther side of the triangle.

SVppose that there be a triangle ABC, vnto one of the sides whereof, namely, vnto BC, let there be drawen a parallel line DE cuttyng the sides AC and AB in the pointes E and D. Then I say first that as BD is to DA, so is CE to EA.* 1.20 Draw a line from B to E, & also from C to D. Wher¦fore (by the 37. of the first) the triangle BDE is equall vnto the triangle CDE: for they are set vpon one and the same base DE, and are contained within the selfe same parallels DE and BC. Consider

[illustration]

also a certaine other triangle ADE. Now thinges equall (by the 7. of the fifth) haue to one selfe thing one and the same proportion. Wherfore as y^e triangle BDE is to y^e triangle ADE, so is y^e triangle CDE to the triangle ADE. But as y^e triangle BDE is to y^e trian∣gle ADE, so is y^e base BD to y^e base DA (by the first of this booke.) For they are vnder one and the selfe same toppe, namely, E, and ther∣fore are vnder one and the same altitude. And by the same reason as the triangle CDE is to the triangle ADE, so is the lyne CE to the lyne EA. Wherfore (by the 11. of the fifth) as the line BD is to the line DA, so is the line CE to the line EA.

But now suppose that in y^e triangle ABC the sides AB & AC be cut propor∣tionally so y^t as BD is to DA, so let CE be to EA, & draw a line from D to E. Then secondly I say y^t the line DE is a parallel to y^e lyne BC.* 1.21 For the same order of construction being kept, for y^t as BD is to DA, so is CE to EA, but as BD is to DA, so is y^e triangle BDE to y^e triangle ADE (by the 1. of the sixt) & as CE is to EA, so (by y^e same) is the triangle CDE to y^e triangle ADE: therfore (by the 11. of the fifth) as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. Wherfore either of these triangles BDE and CDE haue to the triangle ADE one and the same proportion. Wherefore (by the 9. of the fifth) the triangle BDE is equall vnto the triangle CDE, and they are vpon one and the selfe base, namely, DE. But triangles equall and set vpon one base, are also contained within the same parallel lines (by the 39. of the first.) Wherfore the line DE is vnto the line BC a parallel. If therfore to a∣ny one of the sides of a triangle be drawn a parallel line, it cutteth the other sides of the same triangle proportionally. And if the sides of a triangle be cut propor∣tionally, a right lyne drawen from section to section, is parallel to the other side of the triangle: which thing was required to be demonstrated.

¶ Here also Flussates addeth a Corollary.

If a line parallel to one of the sides of a triangle do cut the triangle, it shall cut of from the whole triangle a triangle like to the whole triangle.* 1.22 For as it hath bene proued it deui∣deth the sides proportionally. So that as EC is to EA, so is BD to DA, wherfore by the 18. of the fifth, as AC is to AE, so is AB to AD. Wherfore alternately by the 16. of the fifth as AC is to AB, so is AE to AD: wherefore in the two trian∣gles EAD and CAB the sides about the common angle A are proportional. The sayd triangles also are equiangle. For forasmuch as the right lynes AEC and ADB do fall vpon the parallel lynes ED and CB, therefore by the 29. of the firs•• they make the angles AED and ADE in the triangle ADE equall to the angles ACB and ABC in the triangle ACB. Wherefore by the first definition of this booke the whole triangle ABC is like vnto the triangle cut of ADE.

The 3. Theoreme. The 3. Proposition. If an angle of a triangle be deuided into two equall partes, and if the right line which deuideth the angle deuide also the base: the segmentes of the base shall be in the same proportion the one to the other, that the other sides of the triangle are. And if the segmētes of the base be in the same proportion that the other sides of the sayd triangle are: a right drawen from the toppe of the triangle vnto the section, shall deuide the an∣gle of the triangle into two equall partes.

SVppose that there be a triangle ABC, and (by the 9. of the first) let the angle BAC be deuided into two equall partes by the right lyne AD, which let cut also the base BC in the point D. Then I say that as the segment BD is to y^e segment DC, so is the side BA to the side AC.* 1.23 For by the point C (by the 31. of the first) draw

[illustration]

vnto the line DA a parallel line CE and extende the line BA till it concurre with the line CE in the point E, and do make the triangle BEC.* 1.24 But the lyne BA shall concurre with the line CE (by the 5. peticion) for that the angles EBC, and BCE are lesse then two right angles. For the angle ECB is equall to the outwarde and opposite angle ADB (by the 29. of the first.) And the two angles ADB and DBA of the triangle BAD are lesse then two right angles (by the 17. of the first) Now for∣asmuch as vpon the parallels AD and EC falleth the right line AC, therefore

by the 29. of the first) the angle ACE is equall vnto the angle CAD. But vn∣to the angle CAD is the angle BAD supposed to be equall. Wherfore the an∣gle BAD is also equall vnto the angle ACE. Againe because vpon the paral∣lels AD and EC falleth the right line BAE, the outward angle BAD (by the 28. of the first) is equall vnto the inward angle AEC. But before it was pro∣uell that y^e angle ACE is equall vnto y^e angle BAD, wherfore y^e angle ACE is equall vnto y^e angle AEC. Wherefore (by y^e 6. of y^e first) y^e side AE is equall vnto the side AC. And because to one of y^e sides of y^e triangle BCE, namely, to EC is drawen a parallel line AD, therfore (by y^e 2. of y^e sixt) as BD is to DC, so is BA to AE. But AE is equall vnto AC, therfore as BD is to DC, so is BA to AC. But now suppose that as the segment BD is to the

[illustration]

segment DC,* 1.25 so is the side BA to the side AC, & draw a line from A to D. Then I say that the an∣gle BAC is by y^e right line AD deuided into two equall partes. For the same order of construction re∣mayning, for that as BD is to DC, so is BA to AC, but as BD is to DC, so is BA to AE (by the 2. of the sixt) for vnto one of the sides of the triangle BCE, namely, vnto the side EC is drawn a parallel line AD. Wherefore also as BA is to AC, so is BA to AE (by the 11. of the fifth) Wher∣fore (by the 9. of the fifth) AC is equall vnto AE. Wherfore also (by the 5. of the first) the angle AEC is equall vnto the angle ACE, but the angle AEC (by the 29. of the first) is equall vnto the outward angle BAD: and the angle ACE is equall vnto the angle CAD which is alternate vnto him: wherefore the angle BAD is equall vnto the angle CAD. Wherfore the angle BAC is by the right line AD deuided into two equall partes. Wherefore if an angle of a triangle be deuided into two equall partes, and if the right line which deuideth the angle cut also the base, the segmentes of the base shall be in the same propor∣tion the one to the other, that the other sides of the said triangle are. And if the segmentes of the base be in the same proportion that the other sides of the sayd triangle are, a right line drawen from the toppe of the triangle vnto the section deuideth the angle of the triangle into two equall partes.

This construction is the halfe part of that Gnomical figure described in the 43. proposition of the first booke, which Gnomical figure is of great vse in a maner in all Geometrical demonstrations.

The 4. Theoreme. The 4. Proposition. In equiangle triangles, the sides which cōtaine the equall an∣gles are proportionall, and the sides which are subtended vn∣der the equall angles are of like proportion.

SVppose that there be two equiangle triangles ABC and DCE: and let the angle ABC of the one triangle, be equall vnto y^e angle DCE of the other triangle, and the angle BAC equall vnto y^e angle CDE, and moreouer, the angle ACB equall vnto the angle DEC. Then I say, that those sides of y^e triangles ABC, & DCE, which include the equall angles, are proportionall, and the side which are subtended vnder the equall an∣gles are of like proportion.* 1.26 For let two sides of the sayd triangles, namely, two of those sides which are subtended vnder equall angles: as for example the sides BC and CE, be so set that they both make one right line. And because the an∣gles ABC & ACB are lesse then two right angles (by the 17. of the first): but the angle ACB is e∣quall

[illustration]

vnto the angle DEC: therfore y^e an∣gles ABC & DEC are lesse thē two right angles. Wherefore the lines BA & ED be∣ing produced, will at the length meete toge∣ther. Let them meete and ioyne together in the poynt F.* 1.27 And because by supposition the angle DCE is equall vnto the an∣gle ABC, therfore the line BF is (by the 28. of the first) a parallell vnto t••e line CD. And forasmuch as by supposition the angle ACB is equall vnto the angle DEC, therefore againe (by the 28. of the first) the line AC is a parallell vnto the line FE. Wherefore FADC is a parallelogramme. Wherfore the side FA is equall vnto the side DC: and the side AC vnto the side FD (by the 34. of the first). And because vnto one of the sides of the triangle BFE, namely, to FE is drawen a parallell line AC, therefore as BA is to AF, so is BC to CE (by the 2. of the sixt). But AF is equall vnto CD. Wherfore (by the 11. of the fift) as BA is to CD, so is BC to CE, which are sides subtended vnder equall angles. Wherefore alternately (by the 16. of the fift) as AB is to BC, so is DC to CE. Againe forasmuch as CD is a parallell vnto BF, therefore againe (by the 2. of the sixt) as BC is to CE, so is FD to DE. But FD is equall vnto AC. Wherefore as BC is, to CE, so is AC to DE, which are also sides sub∣tended vnder equall angles. Wherfore alternately (by the 16. of the fift) ••s BC is to CA, so is CE to ED•• Wherfore forasmuch as it hath bene demonstrated, that as AB is vnto BC•• so is DC vnto CE•• but as DC is vnto CA, so is CE vnto ED•• it followeth of equalitie (by the 22. of the fift) that ••s BA is vnto AC so is CD vnto DE•• Wherfore in eq••iangle triangle••, y^e sides which include y^e equall angles are proportionall: and y^e sides which are subt••nded vnder the equall angles are of like proportion ••hich was required to be demonstrated.

The 5. Theoreme. The 5. Proposition. If two triangles haue their sides proportionall, the triang••••s are equiangle, and those angles in thē are equall, vnder which are subtended sides of like proportion.

SVppose that there be two triangles ABC, & DEF, hauing their sides proportionall, as AB is to BC, so let DE be to EF:* 1.28 & as BC is to AC, so let EF be to DF: and moreouer, as BA is to AC, so let ED be to DF. Then I say, that the triangle ABC is equiangle vnto the triangle DEF: and those angles in them are equall vnder which are subtended sides of like pro∣portion, that is, the angle ABC is equall vnto the angle DEF: and the angle BCA vnto the angle EFD: and moreouer, the angle BAC to y^e angle EDF. Vpon the right line EF,* 1.29 and vnto the pointes in it E & F, describe (by the 23. of the first) angles equall vnto the angles ABC & ACB, which let be FEG and EFG, namely, let the angle FEG be equall vnto the angle ABC, and let the angle EFG be equall to the angle ACB.* 1.30 And forasmuch as the angles ABC and ACB are lesse then two right angles (by the 17. of the first): therefore also the angles FEG and EFG are lesse then two right angles. Wherefore (by the 5. petition of y^e first) y^e right lines EG & FG shall at y^e length concurre. Let thē concurre in the poynt G. Wherefore EFG is a triangle. Wherefore the angle remayning BAC is equall vnto the angle remay∣ning

[illustration]

EGF (by the first Corollary of the 32. of the first). Wherfore the triangle ABC is equiangle vn∣to the triangle GEF. Wherefore in the triangles ABC and EGF the sides, which include the equall angles (by the 4. of the sixt) are proportionall, and the sides which are subtended vnder the equall an∣gles are of like proportion. Wherefore as AB is to BC, so is GE to EF. But as AB is to BC, so by supposition is DE to EF. Wherefore as DE is to EF, so is GE to EF (by the 11. of the fift). Where∣fore either of these DE and EG haue to EF one and the same proportion. Wherefore (by the 9. of the fift) DE is equall vnto EG. And by the same rea∣son also DF is equall vnto FG. Now forasmuch as DE is equall to EG and EF is common vnto them both, therefore these two sides DE & EF, are equall vnto these two sides GE and EF, and y^e base DF is equall vnto the base FG. Wherefore the angle DEF (by the 8. of the first) is equall vnto the angle GEF: and the triangle DEF (by the 4. of the first) is equall vnto the triangle GEF: and the rest of the angles of the one triangle

are equall vnto the rest of the angles of the other triangle the one to y^e other, vn∣der which are subtended equall sides. Wherefore the angle DFE is equall vn∣to the angle GFE: and the angle EDF vnto the angle EGF. And because•• the angle FED is equall vnto the angle GEF: but the angle GEF is equall vnto the angle ABC: therefore the angle ABC is also equall vnto the angle FED. And by the same reason the angle ACB is equall vnto y^e angle DFE•• and moreouer, the angle BAC vnto the angle EDF. Wherefore the triangle ABC is equiangle vnto the triangle DEF. If two triangles therefore haue their sides proportionall, the triangles shall be equiangle, & those angles in them shall be equall, vnder which are subtended sides of like proportion: which was required to be demonstrated.

The 6. Theoreme. The 6. Proposition. If there be two triangles wherof the one hath one angle equall to one angle of the other, & the sides including the equall an∣gles be proportionall: the triangles shall be equiangle, and those angles in them shall be equall, vnder which are subten∣ded sides of like proportion.

SVppose that there be two triangles ABC, and DEF, which let haue the angle BAC of the one triangle equall vnto the angle EDF of the other triangle, and let the sides including the equall angles be proportio∣nall, that is, as BA is to AC, so let ED be to DF. Then I say, that y^e triangle ABC is equiangle vnto the triangle DEF: and the angle ABC is equall vn∣to the angle DEF, and the angle ACB equall vnto the angle DFE, which ••ngles are subtēded to sides of like pro∣portion.

[illustration]

Vnto the right line DF, and to the poynt in it D (by the 23. of the first) describe vnto either of y^e angles BAC and EDF,* 1.31 an equall angle FDG. And vnto the right line DF, and vnto the point in it F (by y^e same) describe vnto y^e angle AC•• an equall angle DFG. And forasmuch as the two angles BAC and ACB, are (by the 17. of the first) lesse then two right angles: therefore also the two angles EDG and DFG, are lesse then two right angles. Wherfore y^e lines DG & FG being produced, shall cōcurre (by the 5. petition). Let thē concurre in the point G. Wherefore DFG is a triangle.

Wherefore the angle remaining ABC is equall vnto the angle remaining DGF (by the 32. of the first). Wherefore the triangle ABC is equiangle vnto the tri∣angle DGF. Wherefore as BA is in proportion to AC, so is GD to DF (by the 4. of the sixt). But it is supposed, that as BA is to AC, so is ED to DF. Wherefore (by the 11. of the fift) as ED is to DF, so is GD to DF. Where∣fore (by the 9. of the fift) ED is equall vnto DG. And DF is common vnto them both. Now then there are two

[illustration]

sides ED and DF equall vnto two sides GD and DF: and the angle EDF (by supposition) is equall vnto the angle GDF. Wherefore (by the 4. of the first) the base EF is equall vnto the base GF, and the triangle DEF is (by the same) equall vnto the triangle GDF, and the other an∣gles remayning in them are equall the one to the other, vnder which are sub∣tended equall sides. Wherefore the an∣gle DFG is equall vnto the angle DFE: and the angle DGF vnto the angle DEF. But the angle DFG is (by construction) equall vnto y^e angle ACB. And the angle DGF is as it hath bene proued, equall to y^e angle ABC. Wherfore also y^e angle ACB is equall vnto the angle DFE. And y^e angle ABC is equall to the angle DEF. But by sup∣position the angle BAC is equall vnto the angle EDF. Wherefore the tri∣angle ABC is equiangle vnto the triangle DEF. If therefore there be two triangles, whereof the one hath one angle equall to one angle of the other, and if also the sides including the equall angles be proportionall: then shall the triangles also be equiangle, and those angles in them shalbe equall, vnder which are subtended sides of like proportion: which was required to be proued.

The 7. Theoreme. The 7. Proposition. If there be two triāgles, wherof the one hath one angle equal to one angle of the other, and the sides which include the other angles, be proportionall, and if either of the other angles re∣mayning be either lesse or not lesse then a right angle: thē shal the triangles be equiangle, and those angles in them shall be equall, which are contayned vnder the sides proportionall.

SVppose that there be two triangles ABC and DEF, which let haue one angle of the one, equall to one angle of the other, namely, the angle BAC equall vnto the angle EDF. And let the sides which include the other angles, namely, the angles ABC and DEF be proportio∣nall, so that as AB is to BC, so let DE be to EF. And let the other angles re∣mayning, namely, ACD and DFE be first either of them lesse then a right angle. Then I say that the triangle ABC is equiangle vnto the triangle DEF. And that the angle ABC is equall vnto the an∣gle

[illustration]

DEF,* 1.32 namely, the angles which are contai∣ned vnder the sides proportionall, and that the angle remayning, namely, y^e angle C is equall vn∣to the angle remayning, namely, to y^e angle F. For first the angle ABC is either equall to the an∣gle DEF, or els vnequall. If the angle ABC be equall to the angle DEF,* 1.33 then the angle remai∣ning, namely, ACB, shall be equall to the angle remayning DFE (by the corollary of the 32. of the first) And therfore the triangles ABC and DEF are equiangle. But if the angle ABC be vnequall vnto the angle DEF, then is the one of them grea¦ter then the other. Let the angle ABC be the greater, and vnto the right line AB and vnto the point in it B (by the 23. of the first) describe vnto the angle DEF an equall angle ABG. And forasmuch as the angle A is equall vnto the angle D, and the angle ABG is equall vnto the angle DEF, therfore the an∣gle remayning AGB is equall vnto the angle remayning DFE (by the corol∣lary of the 32. of the first. Wherfore the triangle ABG is equiangle vnto the tri∣angle DEF. Wherfore (by the 4. of the sixth) as the side AB is to the side BG, so is the side DE to the side EF. But by suppposition the side DE is to the side EF, as the side AB is to the side BC. Wherfore (by the 11. of the fifth) as the side AB is to the side BC so is the same side AB to the side BG. Wher∣fore AB hath to either of these BC and BG one and the same proportion, and therfore (by the 9. of the fifth) BC is equall vnto BG. Wherefore (by the 5. of the first) y^e angle BGC is equall vnto y^e angle BCG: but by supposition y^e angle BCG is lesse then a right angle. Wherfore the angle BGC is also lesse then a right angle. Wherfore (by the 13. of the first) the side angle vnto it, namely, AGB is greater then a right angle, and it is already proued that the same angle is equall vnto the angle F. Wherfore the angle F is also greater then a right angle. But it is supposed to be lesse which is absurde. Wherefore the angle ABC is not vnequall vnto the angle DEF, wherfore it is equall vnto it. And the angle A is equall vnto the angle D by supposition. Wherfore the angle remayning, name∣ly, C, is equall vnto the angle remayning, namely, to F (by the corollary of the 32. of the first) Wherfore the triangle ABC is equiangle vnto the triangle DEF.* 1.34

But now suppose that either of the angles ACB and DFE be not lesse then

a right angle. That is, let either of them be a right angle, or either of them grea∣ter then a right angle. Then I say againe that in that case also the triangle ABC is equiangle vnto the triangle DEF. For if either of them be a right angle, forasmuch as all right angles are (by the 4. peticion) equall the one to the other, straight way will follow the intent of the proposition. But if either of them be greater then a right angle, then the same order of construction that was before being kept, we may in like sort proue that the side BC is equall vnto the side BG. Wherfore also the angle BCG is equall vnto the angle BGC. But the angle BCG is greater then a right angle. Wherfore also the angle BGC is greater thē a right angle. Wherfore two angles of the triangle BGC are greater then two right angles: which (by the 17. of the first) is impossible. Wherfore the angle ABC is not vnequall vnto the angle DEF. And therfore is it equall: but the angle A is equall vnto the angle D (by supposition) Wher∣fore

[illustration]

the angle remayning, namely, C is equal vnto the angle remayning, namely, to F (by the corollary of the 32. of the first). Wherfore the triangle ABC is equi∣angle vnto the triangle DEF. If there••ore there be two triangles whero•• the one hath one angle equall to one angle o•• the other, and the sides which include the other angle be proportionall, and if either of the other angles remaining be either lesse or not lesse then a right angle, the triangle shall be equiangle, and those angles in them shall be equall which are contained vnder sides proportionall: which was required to be proued.

The 8. Theoreme. The 8. Proposition. If in a rectangle triangle be drawen from the right angle vn∣to the base a perpendicular line, the perpendicular line shall deuide the triangle into two triangles like vnto the whole, and also like the one to the other.

SVppose that there be a rectangle triangle ABC, whose right angle let be BAC: and (by the 12. of the first) from the point A to the line BC let there be drawen a perpendicular line AD,* 1.35 which perpendi∣cular line let deuide the whole triangle ABC into these two trian∣gles ABD and ADC. (Note that this perpendicular line AD, drawen from the right angle to the base, must needes fall within the triangle ABC, & so deuide the triangle into two triangles. For if it should fall without, then pro∣ducing the side BC vnto the perpendicular line, there should be made a triangle, whose outward angle being an acute angle, should be lesse then the inward and opposite angle which is a right angle: which is cōtrary to the 16. of the first. Nei∣ther

can it fall vpon any of the sides AB or AC for th••n two angles of one and the selfe same triangle should not be lesse than ••wo right angles, contrary to the selfe same 17. of the first. Wherefore it fall••••h within the triangle ABC)••

Then I say, that either of these trian∣gles

[illustration]

ABD and ADC, are like vnto the whole triangle ABC, and more∣ouer, that they are like the one to the o∣ther. First that the triangle ABD is like vnto the whole triangle ABC is thus proued.* 1.36 Forasmuch as (by the 4. petition) the angle BAC is equall vn∣to the angle ADB, for either of them is a right angle. And in the two triangles ABC and ABD the angle B is common. Wherefore the angle remayning, namely, ACB is (by the Corollary of the 32. of the first) equall vnto the angle remayning, namely, to BAD. Wher∣fore the triangle ABC is equiangle vnto the triangle ABD. Wherefore the sides which containe the equall angles, are (by the 4. of the sixt) proportionall. Wherfore as the side CB•• which subtendeth y^e right angle of the triangle ABC, is vnto the side BA which subtendeth the right angle of the triangle ABD, so is the same side AB which subtendeth the angle C of the triangle ABC, vn∣to the side BD which subtendeth y^e angle BAD of the triangle ABD, which is equall vnto the angle C: and moreouer, the side AC vnto the side AD which subtend the angle B common to both the triangles. Wherfore the triangle ABC is like vnto the triangle ABD (by the 1. definition of the sixt). In like maner also may we proue, that the triangle ADC is like vnto the triangle ABC. For the right angle ADC is equall to the right angle BAC, & the angle at the point C is common to either of those triangles. Wherefore the angle remayning, namely, DAC is equall to the angle remaining•• namely, to ABC (by y^e Corolla∣ry of the 32. of the first). Wherefore the triangles ABC & ADC are equian∣gle. And therefore (by the 4. of the sixt) the sides which are about the equall an∣gles are proportionall. Wherefore as in the triangle ABC the side BC is to the side CA, so in the triangle ADC is the side AC to the side DC: and againe, as in the triangle ABC the side CA is to the side AB, so in the triangle ADC is the side CD to the side DA. And moreouer, as in the triangle ABC the side CB is to the side BA, so in the triangle ADC is the side CA to the side AD. Wherefore the triangle ADC is like vnto the whole triangle ABC. Wherfore either of these triangles ABD, & ADC is like vnto y^e whole triangle ABC.

I say also, that the triangles ABD and ADC are like the one to the other. For forasmuch as the right angle BDA is equall vnto y^e right angle ADC (by the 4. petition) and as it hath already bene proued, the angle BAD is equall vn∣to the angle C: therefore the angle remayning, namely, B is equall vnto the an∣gle remayning, namely, to DAC (by the Corollary of the 3••. of the first). Wher∣fore

the triangle ABD is equiangle vnto the triangle ADC. Wherefore as the side BD which in the triangle ABD subtendeth the angle BAD is vnto the side DA which in the triangle ADC subtendeth the angle C which is e∣quall vnto the angle BAD, so is the side AD which in y^e triangle ABD sub∣tendeth the angle B, vnto the side DC which in the triangle ADC subtendeth the angle DAC which is equall vnto the angle B: and moreouer, so is the side BA vnto the side AC which subtende the right angles. Wherefore the triangle ABD is like vnto the triangle ADC. If therefore in a rectangle triangle be drawen from the right angle vnto the base a perpendicular line, the perpendicu∣lar line shall deuide the triangle into two triangles like vnto the whole, and also like the one to the other: which was required to be proued.

Corollary.

Hereby it is manifest, that if in a rectangle triangle be draw∣en from the right angle vnto the base a perpendicular line, the same line drawen is a meane proportionall betwene the secti∣ons of the base: and moreouer, betwene the whole base and ei∣ther of the sections, the side annext to the sayd section is the meane proportionall. For is was proued, that as CD is to DA, so is DA to DB: and moreouer, as CB is to BA, so is BA to BD: and finally, as BC is to CA, so is CA to CD.

The 1. Probleme. The 9. Proposition. A right line being geuen, to cut of frō it any part appointed.

LEt the right line geuen be AB. It is requi∣red

[illustration]

that from the same line AB be cut of any part appoynted. Suppose that a thyrd part be appoynted to be cut of.* 1.37 From y^e point A draw a right line AC making with the line AB an angle: and in the line AC take a poynt at all aduentures, which let be D. And beginning at D put vnto AD two equall lines DE & EC (by the 2. of the first). And draw a right line from B to C: and by the point D (by the ••1. of the first) draw vnto BC a pa∣rallell line DF.* 1.38 Now forasmuch as vnto one of y^• sides of the triangle ABC, namely, vnto y^e side BC is draw∣en a parallell line FD, it followeth by y^e 2. of this booke,

that as CD is in proportion vnto DA, so is BF to FA. But (by construction) CD is double to DA. Wherefore y^• line BF is also double to the line FA. Wher∣fore the line BA is treble vnto the line AF. Wherfore from the right line•• geuen AB, is cut of a third part appoynted, namely, AF: which was required to be done.

The 2. Probleme. The 10. Proposition. To deuide a right line geuē not deuided, like vnto a right line geuen beyng deuided.

SVppose that the right line geuen not deuided be AB, and the right lyne geuen being deuided, let be AC. It is required to deuide y^e line AB which is not deuided like vnto the line AC which is deuided.* 1.39 Suppose the lyne AC be deuided in the pointes D and E, & let y^• lines AB & AC so be put, that they make an angle at all aduentures, and draw a line from B to C, and by the pointes D and E draw vnto the line BC (by the 31. of the first) two parallel lines DF and EG: and by the point D vnto

[illustration]

the line AB (by the same) draw a parallel line DHK. Wherfore either of these figure•• FH and HB are parallelogrammes. Wherfore the line DH is equall vnto the line FG, and the line HK is equall vnto the line GB.* 1.40 And be∣cause to one of the sides of the triangle DKC, namely, to the side KC is drawn a paral∣lel line HE, therefore the line CE (by th•• 2. of the sixt) is in proportion vnto the line ED as the line KH is to the line HD: but the line KH is equall vnto the line BG, and the line HD is equal vnto the line GF. Wher¦fore (by the 11. of the fift) as CE is vnto ED, so is BG to GF. Agayne because to one of the sides of the triangle AGE, namely, to GE is drawn a parallel lyne FD, therfore the line ED (by the 2. of the sixth) is in proportion vnto the lyne DA, as the line GF is to the line FA. And it is already proued that as CE is to ED, so is BG to GF. VVherfore as CE is to ED, so is BG to GF, and as ED is to DA, so is GF to FA. VVherfore the right line geuen not deuided, name¦ly, AB is deuided like vnto the right line geuen being deuided, which is AC: which was required to be done.

¶ A Corollary out of Flussates.* 1.41

By this Proposition we may deuide any right line geuen, accordyng to the pro∣portion

of any right lynes geuen. For let those right lynes hauyng proportion be ioyned together directly, that they may make all one right lyne, and then ioyne them to the lyne geuen anglewise. And so proceede as in the proposition, where you see that the right line geuen AB is deuided into the right lynes AF, FG and GB which haue the selfe same proportion that the right lines AD, DE, and EC haue.

By this and the former proposition also may a right line geuen be easily deui∣ded into what partes so euer you will name.* 1.42 As if you will deuide the line AB in∣to three equall partes, let the lyne DE be made equall to the lyne AD, and the lyne EC made equall to the same by the third of the first. And then vsing the selfe same maner of construction that was before: the lyne AB shall be deuided into three equall partes. And so of any kynde of partes whatsoeuer.

The 3. Probleme. The 11. Proposition. Vnto two right lines geuen, to finde a third in proportion with them.

SVppose that there be two right lines geuen BA and AC, and let them be so put that they comprehend an angle howsoeuer it be. It is required to finde vnto BA and vnto AC a third line in proportion.* 1.43 Produce y^e lynes AB and AC vnto the pointes D and E. And vnto the

[illustration]

line AC (by the 2. of the first) put an equall line BD, and draw a lyne from B to C. And by the pointe D (by the 31. of the first) draw vnto the lyne BC a parallel lyne DE, which let concurre with the line AC in the point E.* 1.44 Now forasmuch as vnto one of the sides of the triangle ADE, namely, to DE is drawne a parallel line BC: therfore as AB is in proportion vnto BD, so (by the 2. of the sixt) is AC vnto CE. But the lyne BD is equall vnto the line AC. VVherfore as the lyne AB is to the line AC, so is the line AC to the line CE. VVherfore vnto the two right lines geuen AB and AC is found a third line CE in proportiō with them: which was required to be done.

¶ An other way after Pelitarius.

Let the lines AB and BC be set directly in such sort that they both make one right line.* 1.45 Then frō the point A erect the lyne AD makyng with the lyne AB an angle at all aduentures. And put the lyne AD equall to the lyne BC. And draw a right line from D to B which produce beyond the poynt B vnto the point E. And by the point C draw vnto the lyne DA a parallel lyne CE concurring with the lyne DE in the point E. Then I say that the line CE is the third line proportionall with the lines AB and BC. For for••asmuch

as by the 15. of the first, the angle B of the

[illustration]

triangle ABD is equall to the angle B of the tri∣angle CBE, and by the 29. of the same, the angle A is equall to the angle C, and the angle D to the angle E: therefore by the 4. of this booke AB is to DA, as BC is to CE. Wherfore (by the 11. of the fifth) AB is to BC as BC is to CE: which was required to be done.

¶ An other way also after Pelitarius.

Let the lines AB and BC be so ioyned toge∣ther,* 1.46

[illustration]

that they may make a right angle, namely, ABC. And drawe a line from A to C, and from the point C drawe vnto the line AC a perpendicular CD (by the 11. of the first) And produce the lyne CD till it concurre with the line AB produced vn∣to the pointe D. Then I say that the line BD is a third lyne proportionall with the lines AB and BC: which thing is manifest by the corollary of the 8. of this booke.

The 4. Probleme. The 12. Proposition. Vnto three right lines geuen to finde a fourth in proportion with them.

SVppose that the three right lines geuen be A, B, C. It is required to finde vnto A, B, C, a fourth line in proportiō with them. Let there be taken two right lines DE & DF comprehending an angle as it shall happen, namely, EDF.* 1.47 And (by the 2. of the first) vnto the line A put an equall line DG. And vn∣to the line B (by the same)

[illustration]

put an equall line GE. And moreouer, vnto y^e line C put an equall line DH. Then draw a line frō G to H. And by the poynt E (by the 31. of the first) draw vnto the line GH a parallell line EF. Now forasmuch as vnto one of the sides of the triangle DEF,* 1.48 namely, vnto y^• side EF is drawen a parallell line GH: therefore (by the 2. of the sixt) as the line DG is to the line GE, so is the line DH to the line HF.

But the line DG is equall vnto the line A, and the line GE is equall vnto the line B, and the line DH vnto the line C. Wherfore as the line A is vnto the line B, so is the line C vnto the line HF. Wherfore vnto the three right lines geuen A, B, C, is found a fourth line HF in proportion with them: which was requi∣red to be done.

¶ An other way after Campane.

Suppose that there be three right lines AB, BC, and BD. It is required to adde vn∣to them a fourth line in proportion with them.* 1.49 Ioyne AB the first, with BD the third, in such sort that they both make one right line, namely, AD. And vpon the said lyne AB erect from the point B the second

[illustration]

line BC making an angle at all aduen∣tures. And draw a line from A to C. Then by the point D draw the lyne DE parallel to the line AC, which pro∣duce vntill it concurre in the point E, with the line CB being likewise pro∣duced to the point E. Then I say that the line BE is the fourth line in pro∣portion with the lines AB, BC, and BD: so that as AB is to BC, so is BD to BE. For forasmuch as by the 15 and 29. of the first the two triangles ABC and DBE are equiangle, therfore (by the 4. of this booke) AB is to BC, as BD is to BE: which was required to be done.

The 5. Probleme. The 13. Proposition. Vnto two right lines geuen, to finde out a meane proportionall.

SVppose the two right lines geuen to be AB and BC. It is required be∣twene th••se two lines AB and BC to finde out a meane line proportio∣nall. Let the lines AB and BC be so ioyned together that they both make one right line, namely, AC.* 1.50 And vp∣on

[illustration]

the line AC describe a semicircle ADC, and from the poynt B raise vp vnto the line AC (by the 11. of the first) a perpendicular line BD, cut∣ting the circumference in the point D: and draw a line from A to D, and an other from D to C. Now forasmuch as (by the 31. of the third) the angle in the semicircle ADC is a right angle,* 1.51 and for that in the rectangle triangle ADC is drawen from the right angle vnto the base a perpendicular line DB: therefore (by the Corollary of the 8. of the sixt) the line DB is a meane proportionall betwene the segmētes of the base AB & BC. Wherefore betwene the two right lines geuen, AB & BC, is found a meane proportionall DB: which was required to be done.

¶ A Proposition added by Pelitarius.

A meane proportionall beyng geuen, to finde out in a line geuen the two extremes.* 1.52 Now it behoueth that the meane geuen be not greater then the halfe of the lyne geuen.

Suppose that the meane geuen be AB, and let the right line geuen be BC. It is re∣quired in the line BC to finde out two extremes, betwene which AB shal be the mean•• proportionall. So that yet the lyne AB be not greater then the halfe part of the line BC. For so could it not be a meane. Ioyne the lines AB and BC directly in such sort, that they both make one right line, name¦ly,

[illustration]

AC. Then vppon the line BC de∣scribe the semicircle BEC. And from the point A erect vnto the lyne AC a perpendicular line AD: which lyne AD put equal vnto the line AB. And by the point D draw vnto the line AC a parallel line DE, which vndoub∣tedly shall either cut or touch the se∣micircle, as in the point E, for that the line AD is not greater then the semidiameter. Then from the point E draw vnto the line BC a perpendicular line EF (by the 12. of the first) Then I say that the line BC is so deuided in the point F, that the lyne AB is a meane proportionall betwene the lines BF and FC. Which thing is manifest (by the 31. of the third) & corollary of the 8. of this booke. For the line FE is equal to the line AD by the 34. of the first, and so is equall to the line AB: then if we draw the ryght lines BE and CE, there shall be made a rectangle triangle BEC. And so by the sayd co∣rollary, the line BF shall be to the lyne FE (and therfore to the line AB) as the line FE is to the lyne FC: which was required to be done.

Flussates putteth this Proposition added by Pelitarius as a corollary following of this 13. proposition.

The 9. Theoreme. The 14. Proposition. In equall parallelogrammes which haue one angle of the one equall vnto one angle of the other, the sides shall be recipro∣kall, namely, those sides which containe the equall angles. And if parallelogrammes which hauing one angle of the one, equal vnto one angle of the other, haue also their sides reciprokal, namely, those which contayne the equall angles, they shall al∣so be equall.

SVppose that there be two equall Parallelogrammes AB and BC, hauing the angle B of the one equall vnto the angle B of the other.* 1.53 And let the lines DB and DE be set directly in such sort that they both make one right line, namely, DE. And then (by the 14. of the first) shall the lines FB and BG be so set that they shall make also one right line, namely, GF. Then I say, that the sides of the parallelogrammes AB and BC, which containe the equall

angles, are reciprocally proportionall: that is, as BD is to BE, so is GB to BF. Make complete the parallelogramme FE by producing the sides AF and CE, till they concurre in the poynt H. Now forasmuch as the parallelogramme AB is (by supposition) equall vnto the parallelogramme BC, and there is a certaine other parallelogramme FE:* 1.54 therfore (by the 7. of the fift) as the parallelogrāme AB is to the parallelogramme FE, so

[illustration]

is the parallelogramme BC to the pa∣rallelogramme FE. But as the paralle∣logramme AB is to y^e parallelogramme FE, so is the side DB to the side BE (by the first of this booke). And (by the same) as the parallelogramme BC is to the parallelogramme FE, so is the side GB to the side BF. Wherefore also (by the 11. of the fift) as the side DB is to the side BE, so is the side GB to the side BF. Wherefore in the parallelogrammes AB and BC the sides which con∣taine the equall angles, are reciprokally proportionall: which was first required to be proued.

But now suppose that the sides about the equall angles be reciprokally propor∣tionall so that as the side DB is to the side BE,* 1.55 so let the side GB be to the side BF. Then I say, y^t the parallelogramme AB is equall vnto y^e parallelogramme BC. For, for that as the side DB is to the side BE, so is the side GB to the side BF: but as the side DB is to the side BE, so (by the 1. of the sixt) is the paralle∣logramme AB to the parallelogramme FE: and as the side GB is to the side BF, so is the parallelogramme BC to the parallelogramme FE. Wherefore also (by the 11. of the fift) as the parallelogramme AB is to the parallelogrāme FE, so is the parallelogramme BC to the same parallelogramme FE. Wherefore the parallelogramme AB is equall vnto the parallelogramme BC (by the 9. of the fift). Wherefore in equall and equiangle parallelogrammes the sides which con∣taine the equall angles are reciprokall: and if in equiangle parallelogrammes the sides which containe the equall angles be reciprokall, the parallelogrammes also shall be equall: which was required to be proued.

The 10. Theoreme. The 15. Proposition. In equal triangles which haue one angle of the one equall vn∣to one angle of the other , those sides are reciprokal, which in∣clude the equall angles. And those triāgles which hauyng one angle of the one equall vnto one angle of the other, haue also

their sides which include the equall angles reciprokal, are al∣so equall.

SVppose that there be two equall triangles ABC, and ADE hauing one angle of the one equall vnto one angle of the other, namely, the angle BAC equall vnto the angle DAE. Then I say that in those triangles ABC and ADE, the sides which include y^e equal angles, are reciprokallie propor∣tionall,* 1.56 that is, as the side CA is to the side AD, so is the side EA to the side AB. For let y^e lines CA and AD be so put, y^t they both make directly one right line. And so also the lines EA and AB

[illustration]

shal both make one right line (by the 14. of the first) And draw a line from B to D. Now forasmuch as (by supposition) the triangle ABC is equall vnto the triangle ADE.* 1.57 And there is a certaine other tri∣angle BAD, vnto which the two equall triangles being compared, it will follow by the 7. of the fifth, that as the triangle ABC is vnto the triangle BAD, so is y^e trian∣gle EAD to the same triangle BAD. But as the triangle ABC is to the triā∣gle BAD, so by the 1. of the sixth, is the base CA to the base AD: and as the triangle EAD is to the triangle BAD, so (by the same) is the base EA to the base AB. Wherfore (by the 11. of the fifth) as the side CA is to the side AD, so is the side EA to the side AB. Wherefore in the triangles ABC and ADE the sides which include the equall angles are reciprokally proportionall.

But now suppose that in the triangles ABC and ADE, the sides which in∣clude the equall angles, be reciprokally proportionall,* 1.58 so that as the side CA is to the side AD, so let the side EA be to the side AB. Then I say that the triangle ABC is equall vnto the triangle ADE. For agayne draw a line from B to D. And for that as the line CA is to the line AD, so is the line EA to the line AB, but as the line CA is to the line AD, so is the triangle ABC to the triangle BAD, and as the line EA is to the line AB so is the triangle EAD to the tri∣angle BAD. Wherfore as the triangle ABC is to the triangle BAD so is the triangle EAD to y^e same triangle BAD. Wherfore either of these triangles ABC and EAD haue vnto y^• triangle BAD one and y^• selfe same proportion. Wher¦fore (by the 9. of the fifth) the triangle ABC is equal vnto the triangle EAD. If therfore there be taken equall triangles hauyng one angle of the one equal vn∣to one angle of the other, those sides in them shal be reciprokal, which include the equal angles. and those triangles which hauing one angle of the one equall vnto one angle of the other, haue also their sides which include the equall angles reci∣prokal, shal also be equall: which was required to be proued.

The 11. Theoreme. The 16. Proposition. If there be foure right lines in proportion, the rectangle figure comprehended vnder the extremes: is equall to the rectangle figure contayned vnder the meanes. And if the rectangle fi∣gure which is contained vnder the extremes, be equall vnto the rectangle figure which is contayned vnder the meanes: then are those foure lines in proportion.

SVppose that there b•• foure right lines in proportiō, namely, AB, CD, E, and F: so that as the line AB is to the line CD, so let the line E be to the line F. Then I say, that the rectangle figure comprehended vn∣der the extremes AB and F,* 1.59 is equall vnto the rectangle figure con∣tayned vnder the meanes CD and E. From the poynt A (by the 11. of the first) raise vp vnto the right line AB a perpendicular line AG. And (by the same) from the point C vnto the right line CD raise vp a perpendicular line CH. And (by the 2. of the first) put the line AG equall vnto the line F, and put also y^e line CH equall vnto the line E, and make complete the parallelogrammes GB and HD. Now for that by supposition as the line AB is to the line CD, so is the line E to the line F. But the line E is equall vnto the line CH, & the line F vn∣to the line AG, therefore as the line AB is to the line CD, so is the line CH to the line AG. Wherefore in the

[illustration]

parallelogrāmes BG and DH the sides which include y^e equall angles, are reciprokally proportionall. But equiangle parallelogrammes whose sides which include the equall an∣gles, are reciprokall, are also equall (by the 14. of the sixt). Wherefore the parallelogramme BG is equall vnto the parallelogrāme DH. But the parallelogramme BG is that which is contayned vnder the lines AB and F, for the line AG is put equall vnto the line F. And the parallelogramme DH is that which is contained vnder the lines CD and E, for the line CH is put equall vnto the line E. Where¦fore the rectangle figure contained vnder the lines AB and F, is equall vnto the rectangle figure contayned vnder the lines CD and E.* 1.60

But now suppose that the rectangle figure comprehended vnder the lines AB and F, be equall vnto the rectangle figure cōprehended vnder y^e lines CD & E.

Then I say, that the foure right lines AB, CD, E and F, are proportionall, that is, as the line AB is to the line CD, so is the line E to the line F. The same or∣der of construction that was before being kept, forasmuch as that which is con∣tained vnder the lines AB and F is equall vnto that which is contained vnder the lines CD and E, but that which is contayned vnder the lines AB and F is the parallelogramme BG, for the line AG is equall vnto the line F. And that also which is contained vnder the lines CD & E is the parallelogramme DH, for the line CH is equall vnto the line E. Wherefore the parallelogramme BG is equall vnto the parallelogramme DH, & they are also equiangle. But in pa∣rallelogrammes equall & equiangle the sides which include the equall angles are reciprokall (by the 14. of the sixt). Wherfore as the line AB is to the line CD, so is the line CH to the line AG, but the line CH is equall vnto the line E, and the line AG is equall vnto y^e line F. Wherefore as the line AB is to the line CD, so is the line E to the line F. If therefore there be foure right lines in proportion, the rectangle figure comprehended vnder the extremes, is equall to the rectangle figure contayned vnder the meanes. And if the rectangle figure which is contai∣ned vnder the extremes, be equall vnto the rectangle figure which is contained vnder the meanes, then are those foure lines in proportion•• which was required to be proued.

The 12. Theoreme. The 17. Proposition. If there be three right lines in proportion, the rectangle figure comprehended vnder the extremes, is equall vnto the square that is made of the meane. And if the rectangle figure which is made of the extremes, be equal vnto the square made of the meane, then are those three right lines proportional.

SVppose that there be three lines in proportion A, B, C, so that as A is to B, so let B be to C. Then I say that the rectangle figure comprehended vnder y^e lines A and C is equall vnto y^e square made of the line B.* 1.61 Vnto the line B (by the 2. of the

[illustration]

first) put an equall line D. And because by supposition) as A is to B, so is B to C, but B is equall vnto D, where∣fore (by the 7. of the fifth) as A is to B, so is D to C, but if there be foure right lines pro¦portionall, the rectangle fi∣gure comprehended vnder

the extremes is equall vnto the rectangle figure comprehended vnder the meanes (by the 16. of the sixt). Wherfore that which is contained vnder the lines A and C is equall vnto that which is comprehended vnder the lines B and D. But that which is contained vnder the lines B and D, is the square of the line B, for the line B is equall vnto the line D. Wherfore the rectangle figure comprehēded vn∣der the lines A and C is equall vnto the square made of the lyne B.

* 1.62But now suppose that that

[illustration]

which is comprehended vn∣der the lines A & C be equal vnto the square made of the line B. Then also I say, that as the line A is to the line B, so is the line B to the lyne C. The same order of constru∣ction that was before, beyng kept, forasmuch as y^t which is contained vnder the lynes A and C is equall vnto the square which is made of the line B. But the square which is made of the line B is that which is contained vnder y^e lines B & D, for the line B is put equall vnto the line D. Wherefore that which is contayned vn∣der the lines A and C is equall vnto that which is contayned vnder the lines B and D. But if the rectangle figure comprehended vnder the extremes, be equall vnto the rectangle figure comprehended vnder the meane lynes, the foure right lines shall be proportionall (by the 16. of the sixth) Wherfore as the line A is to the line B, so is the line D to the line C. But the line B is equall vnto the lyne D. Wherfore as the line A is to the lyne B, so is B to the line C. If therefore there be three right lynes in proportion, the rectangle figure comprehended vnder the ex∣tremes, is equall vnto the square that is made of the meane. And if the rectangle figure which is contayned vnder the extremes, be equall vnto the square made of the meane, then are those three right lines proportional: which was required to be demonstrated.

* 1.63¶ Corollary added by Flussates.

Hereby we gather that euery right lyne is a meane proportionall betwene euery two right lines which make a rectangle figure equall to the square of the same right lyne.

The 6. Probleme. The 18. Proposition. Vpon a right line geuen, to describe a rectiline figure like, and in like sort situate vnto a rectiline figure geuen.

SVppose that the ri••ht line geuen be AB•• and let the rectiline figure ge∣uen be EG. It is required vpon the right line geuen AB to describe a rectiline figure like, and in like sort situate vnto the rectiline figure ge∣uen GE. Drawe a line from H to F, and vnto the right line AB and to y^e p••i••t in it A,* 1.64 make vnto the angle E an equall angle DAB (by the 23. of the firs••) and vnto the right line AB and vnto the point in it B (by the same) make vn••o the angle EFH an equall angle ABD. Wherefore y^e angle remayning EHF is equall vnto the angle remayning ADB. Wherefore the triangle HEF is equiangle vnto the triangle DAB. Wherefore (by the 4. of the sixt) as the side HF is in proportion

[illustration]

to the side DB, so is the side HE to the side DA, and y^• side EF to the side AB. Againe (by the 23. of the first) vnto the right line BD and vnto y^e point in it D, make vnto the angle FHG an equall an∣gle BDC, and (by the same) vnto the right line BD and vnto the point in 〈◊〉〈◊〉 B, make vnto the angle HFG an ••quall angle DBC.* 1.65 Wherefore the angle remayning, namely, G, is equall vnto the angle remayning, namely, to C. Where∣fore the triangle HFG is equiangle vnto the triangle DBC. Wherefore (by the 4. of the sixt) as the side HF is in proportion to the side DB, so is the side HG to the side DC, and the side GF to the side CB. And it is already proued that as HF is to DB, so is HE to DA, and EF to AB. Wherefore (by the 11. of the fift) as EH is to AD, so is EF to AB, and HG to DC, and more∣ouer, GF to CB. And forasmuch as the angle EHF is equall vnto the angle ADB, and the angle FHG is equall vnto the angle BDC: therefore the whole angle EHG is equall vnto the whole angle ADC, and by the same rea∣son the angle EFG is equall vnto the angle ABC. But (by construction) the angle E is equall vnto the angle A, and the angle G is proued equall vnto the an∣gle C. Wherefore the figure AC is equiangle, vnto the figure EG, and those sides which in it include the equall angles are proportionall, as we haue before proued. Wherefore the rectiline figure AC is (by the first definition of the sixt) like vnto the re••••••line figure geuen EG•• Wherefore vpon the right line geuen AB is described a rectiline figure AC like & in like sort ••••••uat•• v••to the recti∣line figure geuen EG: which was required to be done.

The 13. Theoreme. The 19. Proposition.

Like triangles are one to the other in double proportion that the sides of lyke proportion are.

SVppose the triangles like to be ABC and DEF, hauing the angle B of the one triangle, equal vnto the angle E of the other triangle, & as AB is to BC, so let DE be to EF, so that let BC & EF be sides of like proportion. Then I say that the proportion of the triangle ABC vnto the triangle DEF is double to the proportion of the side BC to the side EF. Vnto the two lines BC and EF (by the 10. of the sixth) make a third lyne in proportion BG, so that as BC is to EF, so let EF be to BG, and draw a lyne from A to G. Now forasmuch as AB is to BC, as DE is to EF, therfore alter∣nately (by the 16. of the fifth) as AB is to DE, so is BC to EF.* 1.66 But as BC is to EF, so is EF to BG, wherfore also (by the

[illustration]

11. of the fifth) as AB is to DE, so is EF to BG. Wherfore the sides of the triangles ABG & DEF, which include y^e equal angles are reciprokally proportionall. But if in triangles hauing one angle of the one equall to one angle of y^e other, the sides which include y^e equall an∣gles, be reciprokal, the triangles also (by y^e 15. the sixth) shal be equall. Wherfore the triangle ABG is equall vnto y^e triangle DEF. And for y^t as y^• line BC is to y^e line EF, so is the line EF to y^e line BG: but if there be three lines in proportion, the first shall haue to the third double proportion that it hath to the second (by the 10. de∣finition of the fifth) therfore the line BC hath vnto the line BG double propor∣tion that it hath to the line EF. But as BC is to BG, so (by the 1. of the sixth) is the triangle ABC to the triangle ABG. Wherfore the tiangle ABC is vnto the triangle ABG in double proportion that the side BC is to the side EF. But the triangle ABG is equall to the triangle DEF. Wherfore also the triangle ABC is vnto the triangle DEF in double proportion that the side BC is to the side EF. Wherfore lyke triangles are one to the other in double proportion that the sides of like proportion are: which was required to be proued.

Corollary.

* 1.67Hereby it is manifest that if there be three right lines in pro∣portion, as the first is to the third, so is the triangle described vpon the first, vnto the triangle described vpon the second, so that the sayd triangles be like, and in lyke ••ort described, for it hath bene proued that as the lyne CB is to the line BG, so

is the triangle ABC to the triangle DEF: which was re∣quired to be demonstrated.

The 14. Theoreme. The 20. Proposition. Like Poligonon figures, are deuided into like triangles and equall in number, and of like proportion to the whole. And the one Poligonon fig••re is to the other Poligonon figure in double proportion that one of the sides of like proportion is to one of the sides of like proportion.

SVppose y^t the like Poligonon figures be ABCDE, & FGHKL, hauing the angle at the point F equall to the angle at the point A, and the angle at the point •• equall to the angle at the point B, and the an∣gle at the point H equall to y^• angle at the point C: and so of the rest. And moreouer, as the side AB is to the side BC, so let the side FG be to the side GH, and as the side BC is to the side CD, so let the side GH be to the side HK and so forth. And let the sides AB & FG be sides of like proportion. Then I

[illustration]

say first,* 1.68 that these Poligonon figures ABCD∣E & FG∣HKL, are deui∣ded into like trian∣gles and equall in number. For draw these right lines, AC, AD, FH, & FK•• And forasmuch as (by supposition, that is, by reason the figure ABCDE is like vnto the figure FGHKL) the angle B is equall vnto the angle G, and as the side AB is to the side BC, so is the side FG to the side GH, it followeth that the two triangles ABC and FGH haue one angle of the one equall to one angle of the other, and haue also the sides about the equall angles proportionall. Wherefore (by the 6. of the sixt) the triangle ABC is equiangle vnto the trian∣gle FGH. And those angles in thē are equall, vnder which are subtended sides of like proportion: namely, the angle BAC is equall to the angle GFH, and the angle BCA to the angle GHF. Wherefore (by the 4. of the sixt) the sides which are about the equall angles are proportionall: and the sides which are sub∣tended

vnder the equall angles are of like proportion. Wherfore as AC is to BC, so is FH to GH. But by supposition as BC is to CD, so is GH to HK. Wherefore of equalitie (by the 22. of the fift) as AC is to CD, so is FH to HK. And forasmuch as by supposition the whole angle BCD is equall to the whole angle GHK, and it is proued that the angle BCA is e••••all to the an∣gle GHF: therefore the angle remayning ACD is equall to the angle remay∣ning FHK (by the 3. common sentence). Wherefore the ••ria••gl••s ACD, and FHK, haue againe one angle of the one equall to one angle of the other, and the sides which are about the equall sides are proportionall. Wherefore (by the ••ame sixt of this booke) the triangles ACD & FHK are equiangle. And (by the 4. of this booke) the sides, which are about the equall angles 〈◊〉〈◊〉 proportio∣nall. And by the same reason may we proue that the triangle AD•• is equian∣gle vnto the triangle FKL. And that the sides which are about the equall an∣gles are proportionall. Wherefore the triangle ABC is like to y^e triangle FGH, and the triangle ACD to the triangle FHK, and also the triangle ADE to the triangle FKL (by the first definition of this sixt booke). Wherfore the Po∣ligonon figures geuen ABCDE, and FGHKL, are deuided into triangles like and equall in number.

* 1.69I say moreouer, that the triangles are the one to the other, and to the whole Poligonon figures proportionall: that is, as the triangle ABC is to the triangle FGH, so is the triangle ACD to the triangle FHK, and y^e triangle ADE to the triangle FKL: and as the triangle ABC is to the triangle FGH, so is the Poligonon figure ABCDE to the Poligonon figure FGHKL. For for∣asmuch as the triangle ABC is like to the triangle FGH, and AC and FH are sides of like proportion, therfore the proportion of the triangle ABC to the triangle

[illustration]

FGH is double to the pro∣portion of the side AC to the 〈◊〉〈◊〉 FH (by y^e for∣mer Pro∣position). And therefore also the proportion of the triangle ACD to the triangle FKH, is double to the proportion that the same side AC hath to the side FH. Where∣fore (by the 11. of the fift) as the triangle ABC is to the triangle FGH, so is the triangle ACD to the triangle FHK. Againe, forasmuch as the triangle ACD is like to the triangle FHK, and the sides AD & FK are of like pro∣portion: therefore the proportion of the triangle ACD to the triangle FHK

is double to the proportion of the side AD to the side FK (by the foresayd 19. of the sixt). And by the same reason the proportion of the triangle ADE to y^e tri∣angle FKL, is double to the proportion of the same side AD to the side FK. Wherfore (by the 11. of the fift) as the triangle ACD is to the triangle FHK, so is the triangle ADE to the triangle FKL. But as the triangle ACD is to the triangle FHK, so is it proued that the triangle ABC is to the triangle FGH. Wherefore also (by the 11. of the fift) as the triangle ABC is to the tri∣angle FGH, so is the triangle ADE to the triangle FKL. Wherefore the foresayd triangles are proportionall: namely, as ABC is to FGH, so is ACD to FHK, and ADE to FKL. Wherefore (by the 12. of the fift) as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes. Wherefore as the triangle ABC is to the triangle FGH, so is the Poligonon figure ABCDE to the Poligonon figure FGHKL. Wher∣fore the triangles are proportionall both the one to the other, & also to the whole Poligonon figures.

Lastly I say,* 1.70 that the Poligonon figure ABCDE hath to the Poligonon fi∣gure FGHKL a double proportion to that which the side AB hath to the side FG: which are sides of like proportion. For it is proued, that as the trian∣gle ABC is to the triangle FGH, so is the Poligonon figure ABCDE to the Poligonon figure FGHKL. But the triangle ABC hath to the triangle FGH a double proportion to that which the side AB hath to the side FG (by the former 19. Proposition of this booke): for it is proued, that y^e triangle ABC is like to the triangle FGH. Wherefore the proportion of the Poligonon figure ABCDE to the Poligonon figure FGHKL is double to the proportion of the side AB to the side FG: which are sides of like proportion. Wherefore like Poligonon figures are deuided. &c. as before•• which was required to be proued.

The first Corollary.

Hereby it is manifest, that all like rectiline figures what soe∣uer, are the one to the other in double proportion that the sides of like proportion are.* 1.71 For any like rectiline figures whatsoeuer are by this Proposition deuided into like triangles and equall in number.

The second Corollary.

Hereby also it is manifest, that if there be three right lines proportionall,* 1.72 as the first is to the third, so is the figure descri∣bed vppon the first to the figure described vpon the second, so that the sayd figures be like and in like sort described.

For it is proued, that the proportion of the Poligonon figure ABCDE to the Poligonon figure FGHKL is double to y^e proportion of the side AB to the side FG. And if (by the 11. of the sixt) vnto the lines AB and FG we take a third line in proportion, namely, MN, the first line, namely, AB shall haue vnto the third line, namely, to MN, double proportion that it hath to the second line, namely, to FG (by the 10. definition of the fift). Wherfore as y^e line AB is to the line MN, so is the rectiline figure ABC to the rectiline figure FGH, the sayd rectiline figures being like & in like sort described.

The 15. Theoreme. The 21. Proposition. Rectiline figures which are like vnto one and the same recti∣line figure, are also like the one to the other.

SVppose there be two rectiline figures A and B like vnto the rectiline figure C. Then I say that the figure A is also like vnto the figure B. For forasmuch as the figure A is like vnto the figure C,* 1.73 it is also equi∣angle vnto it (by the conuersion of the first definition of the sixth) & the sides including the equall angles

[illustration]

shall be proportionall. Agayne foras∣much as the figure B is like vnto the figure C, it is also (by the same defini∣tion) equiangle vnto it, and the sides about the equall angles are proportio∣nall. Wherfore both these figures A and B are equiangle vnto the figure C, and the sides about the equall angles are proportionall. Wherfore (by the first common sentence) the figure A is equiangle vnto the figure B, and the sides a∣bout the equall angles are proportionall, wherfore the figure B is like vnto the figure A, which was required to be proued.

The 16. Theoreme. The 22. Proposition. If there be foure right lines proportionall, the rectiline figures also described vpon them beyng lyke, and in like sorte situate, shall be proportional. And if the rectiline figures vppon them described be proportional, those right lynes also shall be pro∣portionall.

SVppose there be foure right lines AB, CD, EF, and GH, and as AB is to CD, so let EF be to GH. And vpon the lines AB and CD (by the 1••. of the sixth) let there be described two rectiline figures KAB, and LCD like the one to the other, and in like sort situate. And vpon the lynes EF and GH (by the same) let there be described also two rectiline figures MF and NH like the one to the other, and in like sorte situate.* 1.74 Then I say that as the ••igure KAB is

[illustration]

to the figure LCD, so is the figure MF to the figure NH. Vnto the lines AB and CD (by the 11. of the sixth) make a third lyne in propor∣tion, namely, O: and vnto the lines EF and GH in like sort make a third lyne in a line proportion, namely, P. And for that as the line AB is to the line CD, so is the line EF to the line GH, but as the line CD is to the line O, so is the line GH to the lyne P. Wherfore of equality (by the 22. of the fifth) as the lyne AB is vnto the line O, so is the lyne EF to the line P. But as the line AB is to the line O, so is the figure KAB to the figure LCD (by the second corollary of the 20. of the sixth). And as the line EF is to the lyne P, so is the figure M F to the figure NH. Where∣fore (by the 11. of the fifth) as the figure KAB is to the figure LCD, so is the fi∣gure M F to the figure NH.

But now suppose that as the figure KAB is to the figure LCD,* 1.75 so is the figure M F to the figure NH, then I say that as the line AB is to the line CD, so is the line EF to the line GH. As the line AB is to the lyne CD, so (by the 1••. of the sixth) let the lyne EF be to the lyne QR, and vpon the lyne QR (by the 18. of the sixth) describe vnto either of these figures MF and NH a like fi∣gure, and in like sort situate SR. Now forasmuch as the lyne AB is to the lyne CD, so is the lyne EF to the line QR, and vpon the lines AB and CD are de∣scribed two figures lyke, and in like sort situate KAB and LCD, and vpon the lines EF and QR are described also two figures like, and in like sort situate MF and SR, therfore as the figure KAB is to the figure LCD, so is the figure MF to the figure SR: wherfore also (by the 11. of the fifth) as the figure MF is to the figure SR, so is the figure MF to the figure NH, wherfore the figure

M F hath to either of these figures NH, and SR one and the same proportion, wherfore by the 9. of the fifth, the figure NH is equal vnto the figure SR. And it is vnto it like, and in like sort situate. * 1.76 But in like and equall rectiline figures beyng in like sort situate, the sides of like proportion on which they are described are equall. Wherfore y^e line GH is equall vnto the line QR. And because as the lyne AB is to the line CD, so is the line EF to the line QR, but the line QR is equall vnto the line GH, therfore as the line AB is to he line CD, so is the line EF to the line GH.

[illustration]

If therefore there be foure right lines proportionall, the rectiline figures al∣so described vpon them beyng like and in lyke sort situate shall be proportionall And if the rectiline figures vpon them described beyng like and in like sort situ∣ate be proportionall, those right lines also shall be proportional: which was required to be proued.

An Assumpt.

And now that in like and equall figures, being in like sort situate, the sides of like proportion are also equall (which thing was before in this proposition taken as graunted) may thus be proued.* 1.77 Suppose y^t the rectiline figures NH and SR be equall and like, and as HG is to GN, so let RQ be to QS, and let GH and QR be sides of like proportion. Then I say that the side RQ is equall vnto the side GH. For if they be vnequall, the one of them is greater then the other, let the side RQ be greater then the side HG. And for that as the line RQ is to the line QS, so is the line HG to the line GN, and alternately also (by the 16. of the fifth) as the line RQ is to the line HG, so is the line QS, to the lyne GN, but the line RQ is greater then the line HG. Wherfore also the line QS is grea¦ter then y^e line GN. Wherefore also y^e figure RS is greater then the figure HN but (by supposition) it is equall vnto it, which is impossible. Wherfore y^e line QR is not greater then y^e line GH. In like sorte also may we proue that it is not lesse then it, wherfore it is equall vnto it: which was required to be proued.

Flussates demonstrateth this second part more briefly,* 1.78 by the first corollary of the ••0. of this boke, thus. Forasmuch as the rectiline figures are by supposition in one and the same proportion, and the same pro∣portion is double to the proportion of the sides AB to CD, and EF to GH (by the foresaid corollary) the proportion also of the sides shall be one and the selfe same (by the 7. common sentence) namely, the line AB shall be vnto the line CD as the line EF is to the line GH.

The 17. Theoreme. The 23. Proposition. Equiangle Parallelogrammes haue the one to the other that proportion which is composed of the sides.

SVppose the equiangle Parallelogrammes to be AC and CF, hauing the angle BCD of the one equall to the angle ECG of the other. Then I say, that the parallelogramme AC is vnto the parallelogramme CF in that proportion which is composed of the proportion of their sides, that is, of that which the side BC hath to the side CG, and of that which the side DC hath to the side CE. Let the lines BC and CG be so put that they both make one right line (by the 14. of the first). Wherefore

[illustration]

(by the same) the lines DC and CE shall make also one right line. Make complete the parallelogramme DG by producing the sides AD and FG till they concurre in the point H, and let there be put a certaine right line K. And as the line BC is to the line CG, so (by the 12. of the sixt) put vnto the line K a line in the same proportion which let be L•• and as DC is to CE, so vnto L put a line in the same pro∣portion, namely, M. Wherefore the proportions of the lines K to L, and L to M, are one and y^e same with the proportions of the sides BC to CG, and D•• to CE: but the proportion of K to M, is composed of the proportions of K to L, & L to M:* 1.79 Wherefore the proportion of K to M, is composed of the proportions of the sides BC to CG, & EC to CD. And for that as the line BC is to the line CG, so is the parallelogramme AC to the parallelogrāme CH (by the 1. of the sixt). But as the line BC is to the line CG, so is the line K to the line L. Wherefore also (by the 11. of the fift) as the line K is to the line L•• so is the parallelogramme AC to the parallelogramme CH. Againe, for that as the line DC is to the line CE, so is the parallelogramme CH to the parallelogramme CF: but as the line DC is to the line CE, so is the line L to the line M: Wherefore also (by y^e same) as the line L is to the line M, so is the parallelogramme CH to the parallelo∣gramme CF. And forasmuch as it is proued, that as the line K is to the line L, so is the parallelogramme AC to the parallelogramme CH, and as the line L is to the line M, so is the parallelogramme CH to the parallelogramme CF: ther∣fore of equalitie (by the 22. of the fift) as the line K is to the line M, so i•• the pa∣rallelogramme AC to the parallelogramme CF. But as it hath before bene pro∣ued, the proportion of the line K to the line M, is composed of the proportions of

the sides BC to CG, and DC to CE. Wherefore also the proportion of the pa∣rallelogramme AC to the parallelogramme CF, is composed of the proportions of the sides BC to CG, and DC to CE. Wherefore equiangle parallelogrammes haue the one to the other that proportion which is composed of the proportions of the sides: which was required to be proued.

Flussates demonstrateth this Theoreme without taking of these three lines, K, L, M, after this maner.

* 1.80Forasmuch as (sayth he) it hath bene declared vpon the 10. definition of the fift booke, and ••ift definition of this booke, that the proportions of the extremes consist of the proportions of the meanes, let vs suppose two equiangle parallelogrāmes ABGD, and GEZI, and let the angles at the poynt G in eyther be equall. And let the lines BG and GI be set directly that they both make one

[illustration]

right line, namely, BGI. Wherefore EGD also shall be one right line by the conuerse of the 15. of the first. Make complete the parallelogramme GT. Then I say, that the proportion of the paral∣lelogrammes AG & GZ is composed of the pro∣portions of the sides BG to GI, and DG to GE. For forasmuch as that there are three magni∣tudes, AG, GT, and GZ, and GT is the meane of the sayd magnitudes: and the proportion of the extremes AG to GZ consisteth of the meane pro∣portions (by the 5. definition of this booke) namely, of the proportion of AG to GT, and of the proportion GT to GZ: But the proportion of AG to GT is one and the selfe same with the proportion of the sides BG to GI (by the first of this booke). And the proportion also of GT to GZ is one and the selfe same with the proportion of the other sides, namely, DG to GE (by the same Proposition). Wherefore the proportion of the parallelogrammes AG to GZ consisteth of the proportions of the sides BG to GI, and DC to GE. Wherefore equiangle parallelogrammes are the one to the other in that proportion which is composed of theyr sides: which was required to be proued.

The 18. Theoreme. The 24. Proposition. In euery parallelogramme, the parallelogrammes about the dimecient are lyke vnto the whole, and also lyke the one to the other.

* 1.81SVppose y^t there be a parallelogramme ABCD, and let the dimecient therof be AC: and let the parallelogrammes about the dimecient AC, be EG and HK. Then I say that either of these parallelogrames EG and HK is like vnto the whole parallelogramme ABCD, and also are lyke the one to the other. For forasmuch as to one of the sides of the trian∣gle ABC, namely, to BC is drawen a parallel lyne EF, therfore as BE is to EA, so (by the 2. of the sixt) is CF to FA. Agayne forasmuch as to one of y^e sides

of the triangle ADC, namely, to CD is drawen a parallel lyne F••, therefore (by the same) as CF is to FA, so is DG to GA. But as CF is to FA, so is it pro••ued that BE is to EA. Whe••fore as BE is to EA, so (by the 11. of the fifth)•• is DG to GA. Wherfore by composition (by the 18. of the fifth) as BA is to AE•• so is DA to AG. And alternately (by the 16. of the fifth) as BA is to AD, so is EA to AG. Wherfore in the parallelogrammes•• ABCD and EG y^e sides which are about the common angle BAD are proportionall. And because y^e line GF is a parallel vnto the lyne DC•• therfore the angle AGF (by the 29•• of the•• first) is equall vnto y^e angle ADC, •• y^e

[illustration]

angle GFA equall vnto y^e angle DCA and the angle DAC is common to the two triangles ADC and AFG: Wher∣fore the triangle DAC is equiangle vnto the triangle AGF. And by the same reason the triangle ABC is equi∣angle vnto the triangle AEF. Wher∣fore the whole parallelogramme ABCD is equiangle vnto the parallelogrāme EG. Wherfore as AD is in proportion to DC, so (by the 4. of the sixth) is AG to GF, and as DC is to CA, so is GF to FA. And as AC is to CB, so is AF to FE. And moreouer as CB is to BA, so is FE to EA. And forasmuch as it is proued that as D•• is to CA, so is GF to FA: but as AC is to C••, so is AF to FE. Wherfore of equalitie (by the 22. of the fifth) as DC is to CB, so is GF to FE. Wherefore in the parallelogrammes ABCD and EG, the sides which include the equall angles are proportionall. Wherefore the parallelogramme ABCD is (by the first definition of the sixth) like vnto the parallelogramme EG.

And by the same reason also the parallelogramme ABCD is like to the pa∣rallelogramme KH:* 1.82 wherefore either of these parallelogrammes EG and KH is like vnto the parallelogramme ABCD. But rectiline figures which are like to one and the same rectiline figure are also (by the 21. of the sixth) like the one to the other. Wherefore the parallelogramme EG is like to the parallelo∣gramme HK.* 1.83 Wherfore in euery parallelogramme, the parallelogrammes a∣bout the dimecient are like vnto the whole, and also like the one to the other. Which was required to be proued.

¶ An other more briefe demonstration after Flussates.

Suppose that there be a parallelogrāme ABCD, whose dime••ient let b•• A••,* 1.84 about which let consist these parallelogrammes EK and TI, hauing the angles at the pointes •• and 〈…〉〈…〉 with the whole parallelogramme ABCD. Then I say, that those pa∣rallelogrammes EK and TI are like to the whole parallelogramme DB and also al••

like the one to the other. For forasmuch as BD, EK,

[illustration]

and TI are parallelogrammes, therefore the right line AZG falling vpon these parallell lines AEB, KZT, and DI G, or vpon these parallell lines AKD, EZI, and BTG, maketh these angles equall the one to the other, namely, the angle EAZ to the angle KZA, & the an∣gle EZA to the angle KAZ, and the angle TZG to the angle ZGI, and the angle TGZ to the angle IZG, and the angle BAG to the angle AGD: and finally, the angle BGA to the angle DAG. Wherefore (by the first Corollary of the 32. of the first, and by the 34. of the first) the angles remayning are equall the one to the other, namely, the angle B to the angle D, and the angle E to the angle K, and the angle T to the angle I. Wherefore these triangles are equiangle and therefore like the one to the other, namely, the triangle ABG to the triangle GDA, and the triangle AEZ to the triangle ZKA, & the triangle ZTG to the triangle GIZ. Wherefore as the side AB is to the side BG, so is the side AE to the side EZ, and the side ZT to the side TG. Wherefore the parallelogrammes contayned vnder those right lines, namely, the parallelogrammes ABGD, EK, & TI, are like the one to the other (by the first definition of this booke). Wherefore in euery parallelogramme the parallelogrammes. &c. as before: which was required to be de∣monstrated.

¶ A Probleme added by Pelitarius.

Two equiangle Parallelogrammes being geuen, so that they be not like, to cut of from one of them a parallelogramme like vnto the other.

* 1.85Suppose that the two equiangle parallelogrammes be ABCD and CEFG, which let not be like the one to the other. It is required from the Parallelogramme ABCD, to cut of a parallelogramme like vnto the parallelogramme CEFG. Let the angle C of the one be equall to the angle C of the other. And let the two parallelogrammes be so 〈◊〉〈◊〉, that the lines BC & CG may make both

[illustration]

one right line, namely, BG. Wherefore also the right lines DC and CE shall both make one right line, namely, DE. And drawe a line from the poynt F to the poynt C, and produce the line FC till it cōcurre with the line AD in the poynt H. And draw the line HK parallell to the line CD (by the 31. of the first). Then I say, that from the paralle∣logramme AC is cut of the parallelogrāme CDHK, like vnto the parallelogrāme EG. Which thing is manifest by thys 24. Propo∣sition. For that both the sayd parallelo∣grammes are described about one & the selfe same dimetient. And to the end it might the more plainly be seene, I haue made complete the Parallelogramme ABGL.

¶ An other Probleme added by Pelitarius.

Betwene two rectiline Superficieces, to finde out a meane superficies proportionall.

* 1.86Suppose that the two superficieces be A and B, betwene which it is required to place a meane superficies proportionall. Reduce the sayd two rectiline figures A and B

vnto two like parallelogrāmes (by the 18. of this booke) or if you thinke good reduce eyther of them to a square, (by the last of the second). And let the said two parallelo∣grammes like the one to the other and equall to the superficieces A and B, be CDEF and FGHK. And let the angles F in either of them be equall, which two angles let be placed in such sort, that the two parallelogrammes ED and HG may be about one and the selfe same dimetient CK (which is done by putting the right lines EF and FG in such sort that they both make one right line, namely,

[illustration]

EG). And make cōplete the parallelogrāme CLK M. Then I say, that either of the supplements FL & FM is a meane proportionall betwene the superficieces CF & FK, that is, betwene the superficieces A and B: name∣ly, as the superficies HG is to the superficies FL, so is the same superficies FL to the superficies ED. For by this 24. Proposition the line HF is to the line FD, as the line GF is to the line FE. But (by the first of this booke) as the line HF is to the line FD, so is the su∣perficies HG to the superficies FL: and as the line GF is to the line FE, so also (by the same) is the superficies FL to the superficies ED. Wherfore (by the 11. of the fift) as the superficies HG is to the superficies FL, so is the same superficies FL to the superficies ED: which was required to be done.

The 7. Probleme. The 25. Proposition. Vnto a rectiline figure geuen to describe an other figure lyke, which shal also be equall vnto an other rectiline figure geuen.

SVppose y^t the rectiline figure geuē, wherunto is required an other to be made like be ABC, and let the other rectiline figure whereunto the same is required to be made, equal be D. Now it is required to describe a rectiline figure like vnto the figure ABC, and equall vnto the fi∣gure D.* 1.87 Vppon the line BC

[illustration]

describe (by the 44. of the first) a parallelogramme BE equall vnto the triangle ABC, and by the same vpon the line CE, describe the paralle∣logramme C M equall vnto the rectiline figure D, and in the said parallelogramme let the angle FCE, be equall vnto the angle CBL. And forasmuch as the angle FCE is by construction equall to the angle CBL,* 1.88 adde the an¦gle BCE common to them both. Wherefore the angles LBC and BCE are e∣quall

vnto the angles BCE and ECF, but the angles LBC and BCE are e∣quall to two right angles (by the 29. of the first) wherfore also the angles BCE and ECF are equall to two right angles. Wherfore the lines BC and CF (by the 14. of the first) make both one right line, namely, BF, and in like sort do the lines LE and EM make both one right line, namely, LM. Then (by the 13. of the sixth) take the meane proportionall betwene the lines BC and CF, which let be GH. And (by the 18. of the sixth) vpon the line GH, let there be described a rectiline figure KHG like vnto the rectiline figure ABC, and in like sorte si∣tuate. And for that as the line BC, is to the line GH, so is the line GH to y^e line CF:* 1.89 but if there be thre right lines proportional, as the first is to the third, so is the figure which is described of the first vnto the figure which is described of the second, the said figures being like and in like sort situate (by the second corrella∣ry of the 20. of the sixth) wherfore as the line BC is to the line CF, so is the tri∣angle ABC to the triangle KGH. But as the line BC is to the lyne CF, so is the parallelogramme BE to the parallelogramme EF (by the 1. of the sixth). Wherfore as the triangle A

[illustration]

BC, is to the triangle KGH so is the parallelogramme BE to the parallelogramme EF. Wherfore alternately also (by the 16. of the fifth) as the triangle ABC is to the pa∣rallelogrāme BE, so is the triangle KGH, to the parallelogramme EF: but the triangle ABC is e∣qual vnto the parallelogrāme BE, wherfore also the trian∣gle KGH is equall vnto the parallelogramme EF: but the parallelogramme FE is equall vnto the rectiline figure D. Wherfore also the rectiline figure KGH is equall vnto the rectiline figure D, and the rectiline figure KGH is by supposition like vnto the recti∣line figure ABC. Wherefore there is described a rectiline figure KGH lyke vnto the rectiline figure geuen ABC, and equall vnto the other rectiline figure geuen D: which was required to be done.

The 19. Theoreme. The 26. Proposition. If from a parallelogramme be taken away a parallelograme like vnto the whole and in like sorte set, hauing also an angle common with it, then is the parallelogramme about one and the selfe same dimecient with the whole.

SVppose that there be a parallelogramme ABCD, and from the paral∣lelogramme ABCD, take away a parallelogramme AF like vnto the parallelogramme ABCD, and in like sort situate, hauing also the an∣gle DAB common with it. Then I say, that the parallelogrammes ABCD and AF are both about one and the self same * 1.90 dimecient AFC, that is, that the dimecient AFC of the whole parallelogramme ABCD passeth by the angle F of the parallelogramme AF, and is common to either of the parallelogrammes. For if AC do not passe by the point F, then if it be possible let it passe by some o∣ther point, as AHC doth. Now then the dimetient AHC shall cut eyther the side GF or the side EF of y^e parallelogramme AF. Let it cut y^e side GF in the point H. And (by the 31. of the first) by the point H let there be drawen to ei∣ther of these lines AD and BC a parallel line HK wherfore GK is a paralle∣logramme, and is about one and the selfe same

[illustration]

dimetient with y^e parallelogramme ABCD. And forasmuch as y^e parallelogrammes ABCD and GK are about one and the self same dimecient, therfore (by the 24. of the sixth) the parallelogramme ABCD is like vnto the parallelogramme GK. Wherfore as the line DA is to the line AB so is the line GA to the line AK (by the conuersion of the first definition of the sixth) And for that the pa∣rallelogrammes ABCD, and EG are (by supposition) like, therfore as the line DA is to the lyne AB so is the line GA to the line AE. Wherfore the line GA hath one and the selfe proportion to either of these lines AK and AE. Wherfore (by the 9. of the fifth) the line AK is e∣quall vnto y^e line AE, namely, y^e lesse to y^e greater, which is impossible. The selfe same inconuenience also will follow, if you put the dimetient AC to cut the side FE. Wherfore AC the dimetie••t of the whole parallelogramme ABCD pas∣seth by the angle and poynt F. And therfore the parallelogramme AEFG is a∣bout one and the selfe same dimetient with the whole parallelogramme ABCD. Wherfore if from a parallelogramme be taken away a parallelograme lyke vnto the whole, and in lyke sorte situate, hauing also an angle common with it, then is that parallelogramme about one and the selfe same dimetient with the whole: which was required to be proued.

¶ An other demonstration after Flussates, which proueth this proposition affirmatiuely.

From the parallelogramme ABGD let there be taken away the parallelogramme AEZK like and in like sorte situate with the whole parallelogramme ABGD, and ha∣uing also the angle A common with the whole parallelogramme.* 1.91 Then I say that both their diameters, namely, AZ and AZG do make one and the selfe same right line. De∣uide the sides AB and B•• into two equall partes in the pointes C and F (by the 10. of

the first.) And drawe a line from C to F.

[illustration]

Wherefore the line CF is a parallel to the right line AG (by the corollary added by Campane after the 29. of the first) Wher∣fore the angles BAG and BCF are equall (by the 29. of the first): but the angle EA∣Z is equall vnto the angle BAG (by reasō the parallelogrammes are supposed to be like) wherefore the same angle EAZ is e∣quall to the angle BCF, namely, the out∣ward angle to the inward and opposite an∣gle. Wherfore (by the 28. of the first) the lines AZ and CF are parallel lynes. Now then the lines AZ and AG being parallels to one and the selfe same lyne, namely, to CF do concurre in the point A. Wherefore they are set directly the one to the other, so that they both make one right line (by that which was added in the ende of the 30. proposition of the first) wherfore the parallelo∣grammes ABGD, and AEZK are about one and the selfe fame dimetient: which was required to be proued.

The 20. Theoreme. The 27. Proposition. Of all parallelogrammes applied to a right line wanting in fi∣gure by parallelogrammes like and in like sort situate to that parallelograme which is described of the halfe line: the grea∣test parallelogramme is that which is described of the halfe line being like vnto the want.

* 1.92LEt there be a right line AB, and (by the 10. of the first) deuide it in two equall partes in the point C. And vnto the right line AB apply a parallelogramme AD wanting in figure by the parallelogramme DB, which let be like and in like sort described vnto the parallelo∣gramme described of halfe the line AB, which is, BC. Then I say, that of all the parallelogrammes which

[illustration]

may be applied vnto the line AB and which wāt in figure by parallelogrāmes like and in like sort situate vnto the pa∣rallelogramme DB the greatest is the parallelogramme AD. For vnto y^e right line AB let there be applied a parallelo∣gramme AF wanting in figure by the parallelogramme FB, which let be like and in like sort situate vnto the paralle∣logramme DB. Then I say, that the parallelogramme AD is greater then the parallelogramme AF. For forasmuch

as the parallelogramme DB is like vnto the parallelogramme ••B, therfore (by the 26. of the sixt) they are about one and the selfe same dimetient.* 1.93 Let their di∣metient be DB, and make complete the figure. Now forasmuch as (by the 43. of y^e first) the supplement FC is equall vnto the supplement FE•• adde the figure FB common to them both. Wherefore the whole figure CR is equall vnto the whole figure KE. But the figure CR is equall vnto the figure CG (by the 36. of the first) for that the base AC is equall vnto the base CB. Wherefore the fi∣gure GC is equall vnto the figure KE. Adde the figure CF common vnto them both. Wherfore the whole figure AF is equall vnto the whole Gnomon LMN. But the whole parallelogramme DB is greater then the Gnomon LMN (by the 3. common sentence). Wherefore also it is greater then the parallelogramme AF. But the parallelogramme AD is equall vnto the parallelogramme DB (by the 36. of the first). Wherefore the parallelogramme AD is greater then the parallelogramme AF. Wherefore of all parallelogrammes applied to a right line wanting in figure by parallelogrammes like and in like sort situate, to that parallelogramme which is described of y^e halfe line, the greatest parallelogrāme is that which is described of the halfe of the line, being like vnto the want: which was required to be proued.

Againe, let AB be deuided into two equall partes in the point C, and let the parallelogramme applied vpon the halfe line be AL,* 1.94 wanting in figure by the parallelogramme LB, which let be like and in like sort situate vnto the paralle∣logramme AL. Againe vnto the line AB let there be applied an other paralle∣logramme AE wanting in figure by y^e pa∣rallelogrāme

[illustration]

EB being like and in like sort situate vnto the parallelogrāme LB which is described vpon halfe of the line AB. Thē I say, that the parallelogramme AL appli∣ed vnto halfe the line is greater then the parallelogramme AE. For forasmuch as the parallelogramme EB is like vnto the parallelogramme LB, they are (by the 26. of the sixt) about one and the same dimeti∣ent. Let their dimetient be EB, and make complete the whole figure, and for that the figure LF is equall vnto the figure LH (by the 36. of the first) for the base FG is equall vnto the base GH, ther∣fore the figure LF is greater then the figure KE. But the figure LF is equall vnto the figure DL (by the 43. of the first). Wherefore the figure DL is greater then the figure KE: put the figure KD common to them both. Wher∣fore the whole parallelogramme AL is greater then the whole parallelogramme AE: which was required to be proued.

The 8. Probleme. The 28. Proposition. Vpon a right line geuen, to apply a parallelogramme equall to a rectiline figure geuen, & wanting in figure by a paralle∣logramme like vnto a parallelogrāme geuen. Now it beho∣ueth that the rectiline figure geuen, whereunto the parallelo∣grāme applied must be equall, be not greater thē that paralle∣logramme, which so is applied vpon the halfe lyne, that the defectes shall be like, namely, the defect of the parallelogrāme applied vpon the halfe line, and the defect of the parallelo∣gramme to be applied (whose defect is required to be like vn∣to the parallelogramme geuen).

SVppose the right line geuen to be AB, and let the rectiline figure ge∣uen wherunto is required to apply vpon the right line AB an equall rectiline figure be C, which figure C, let not be greater then that pa∣rallelogrāme which is so applied vpon the halfe line, that the defectes shall be like, namely, the defect of the parallelogramme applied vpon the halfe line, and the defect of the parallelogramme to be applied (whose defect is requi∣red to be like vnto the parallelogramme geuē). And let the figure whereunto the defect or want of the parallelogramme is

[illustration]

required to be like D. Now it is requi∣red vpō y^e right line geuē AB, to describe vnto the rectiline figure geuen C, an e∣qual parallelogramme wanting in figure by a parallelogramme like vnto D.* 1.95 Let the line AB (by the 10. of the first) be deuided into two equall partes in y^e point E. And (by the 18. of the sixth) vppon the line ••B describe a rectiline figure EBFG like vnto the parallelogramme D and in like sort situate, which shall also be a parallelograme. And make complete the parallelogramme AG.* 1.96 Now then the parallelogramme AG is either equal vnto the rectiline figure C, or greater then it by supposition. If the parallelo∣gramme AG be equal vnto the rectiline figure C, then is that done which we ••ought for.* 1.97 For then vpō the right line AB is described vnto the rectiline figure

geuen C an equal parallelogramme AG wanting in figure by the parallelograme GB,* 1.98 which is like vnto the parallelogramme D. But if AG be not equal vnto C then is AG greater then C, but AG is equall vnto GB (by the first of the sixt). Wherfore also GB is greater then C. Take the excesse of the rectiline figure BG aboue the rectiline figure C (by that which Pelitarius addeth after the 4••. of the first) And vnto that excesse (by the 15. of the sixt) describe an equall recti∣line figure KLMN like and in like sort situate vnto the rectiline figure D. But the rectiline figure D is like vnto the rectiline GB, wherfore also the recti∣line figure KLMN is like vnto the rectiline figure GB (by the 25. of the sixt) Now then let the sides KL and GE be sides of like proportion, let also y^e sides LM and GF be sides of like proportion. And forasmuch as the parallelogrāme GB is equal vnto the figures C and KM, therfore the parallelogramme GB is greater then the parallelogramme KM. Wherefore also the side GE is greater then the side KL, and the side GF is greater then the side LM, vnto the side KL put an equall line GO (by the 2. of the first) and likewise vnto the side LM put an equall line GP. And make perfect the parallelogramme OGPX. Wherfore the parallelogramme GX is equal & like vnto the parallelogramme KM. But the parallelogramme KM is lyke vnto the parallelogramme GB. Wherfore also the parallelogramme GX is like vnto the parallelogramme GB. Wherfore the parallelogrammes GX and GB are (by the 26. of the sixt) about one and the self same dimecient. Let their dimecient be GB, and make complete the figure. Now forasmuch as the parallelogramme BG is equall vnto the recti∣line figure C, and vnto the parallelogramme KM, and the parallelogramme GX, which is part of the parallelogramme GB, is equal vnto KM. Wherfore the Gnomon remayning YQV is equall vnto the rectiline figure remayning, name∣ly, to C. And forasmuch as the supplement PR is equall vnto the supplement OS, put the parallelogramme XB common vnto them both. Wherfore the whole parallelogramme PB is equall vnto the whole parallelograme OB. But the pa∣rallelogramme OB is equal vnto the parallelogramme TE by the 1. of the sixt, (for the side AE is equal vnto the side EB) Wherfore the parallelogramme TE is equal vnto the parallelogramme PB. Put the parallelogramme OS com∣mon to them both. Wherfore the whole parallelogramme TS is equall vnto the whole gnomon YQV. But it is proued that the gnomon YQV is equal vnto the rectiline figure C. Wherfore also the parallelogramme TS is equal vnto the re∣ctiline figure C. Wherfore vpon the right line geuen AB is applied a parallelo∣gramme TS equal vnto the rectiline figure geuen C, and wanting in figure by a parallelogramme XB which is like vnto the parallelogramme geuen D, for the parallelogramme XB is like vnto the parallelogramme GX: which was re∣quired to be done.

¶ A Corollary added by Flussates.

Hereby it is manifest, that if vpon a right line be applied a parallelogramme wantyng

in figure by a square,* 1.99 the parallelogramme ap∣plied

[illustration]

shall be equall to the rectangle figure which is contayned vnder the segments of the line geuen which are made by the application. For the rest of the line is equall to the o∣ther side of the parallelogramme applied. For that they are sides of one & the selfe same square, as the parallelogramme AG is contained vnder the lines AD and DB, or DG which is equall to DB.

The 9. Probleme. The 29. Proposition. Vpon a right line geuen to apply a parallelogramme equall vnto a rectiline figure geuen, and exceeding in figure by a pa∣rallelogramme like vnto a parallelogramme geuen.

SVppose the right line geuen to be AB, and let the rectiline figure geuen whereunto is required vpon the line AB to apply an equall parallelo∣gramme be C: let also the parallelogramme wherunto y^e excesse is requi∣red to be like, be D. Now it is requi∣red

[illustration]

vpon the right line AB to apply in parallelogramme equall vnto the rectiline figure C, and exceeding in fi∣gure by a parallelogramme like vnto the parallelogramme D.* 1.100 Let the line AB be (by the 10. of the first) deui∣ded into two equall partes in the point E. And vpon the line EB (by the 18. of the sixt) describe a parallelo∣gramme BF like vnto the figure D, and in like sort situate. And vnto both

these figures BF and C, describe an equall rectiline figure GH like vnto the figure D and in like sort situate (by the 25. of the sixt). Wherefore the paralle∣logramme GH is (by the 21. of the sixt) like vnto the parallelogramme BF. Let the sides KH and FL be sides of like proportion, and so also let the sides KG and FE be. And forasmuch as the parallelogramme GH is (by constructi∣on) greater then the parallelogramme FB, therefore the line KH is greater then the line FL, and the line KG is greater then the line FE. Extend the lines FL and FE to the pointes M & N, and vnto the line KH put an equall line FLM, and likewise vnto the line KG put an equall line FEN: and make perfect the figure MN. Wherefore the parallelogramme MN is equall and like vnto the parallelogramme GH. But the parallelogramme GH is like vnto the parallelogramme EL. Wherefore also the parallelogramme MN is like vnto the parallelogramme EL. Wherefore the parallelogrammes EL and MN are (by the 26. of the sixt) about one and the same dimetient. Let the sayd dimetient be FO, and make perfect the figure.* 1.101 Now forasmuch as the pa∣rallelogramme GH is equall vnto the figures EL and C. But by construction the parallelogramme GH is equall vnto the parallelogramme MN. Wherfore the parallelogramme MN is equall vnto the figures EL and C. Take away the figure EL which is common to them both. Wherefore the Gnomon remay∣ning, namely, VYX, is equall vnto the rectiline figure C. And forasmuch as the line AE is equall vnto the line EB, therefore the parallelogramme A N is (by the 36. of the first) equall vnto the parallelogramme NB, that is, vnto the parallelogramme LP, which (by the 43. of the first) is equall vnto the pa∣rallelogramme NB. Adde the parallelogramme BO common to them both. Wherefore the whole parallelogramme AO is equall vnto the Gnomon VYX. But the Gnomon VYX is equall vnto the rectiline figure C. Wherefore the pa∣rallelogramme AO is equall vnto the rectiline figure C. Wherefore vpon the right line geuen AB is applied the parallelogramme AO equall vnto the recti∣line figure geuen C, and exceeding in figure by the parallelogramme QP which is like vnto the parallelogramme geuen D. For the parallelogramme D is like vnto the parallelogramme BF: and the parallelogramme BF is like vnto the parallelogramme PQ: for they are about one and the selfe same dimetient: which was required to be done.

The 10. Probleme. The 30. Proposition. To deuide a right line geuen by an extreme and meane pro∣portion.

SVppose the right line geuen to be AB. It is required to devide the line AB by an extreme and meane proportion.* 1.102 Vpon the line AB describe (by the 46. of the first) a square BC. And vpon the line AC (by the

29. of the sixt) applie a parallelogramme CD equall vnto the square BC,* 1.103 and exceding in figure by the figure AD like vnto the figure BC. Now BC is a square. Wherefore also AD is a square. And forasmuch as BC is equall vnto CD, take away the figure CE which is common

[illustration]

to them both. Wherefore the figure remayning, namely, BF, is equall to the figure remayning, namely, to AD, and the angle E of the one is e∣quall vnto the angle E of the other. Wherefore (by the 2. definition of the sixt, and by the 14. of the sixt) the sides of the figures BF and DA, which containe the equall angles, are reciprokall. Wherefore as the side FE is to the side ED, so is the side AE to the side EB. But the side FE is equall vnto the line AC, that is, vnto the line AB, and the side ED is equall vnto the line AE (by the 34. of the first). Wherefore as the line BA is to the line AE, so is the line AE to the line EB. But the line AB is greater then the line AE. Wherfore also the line AE is grea∣ter then the line EB. Wherefore the right line AB is deuided by an extreme and meane proportion in the point E: and the greater segment thereof is AE: which was required to be done.

An other way.

* 1.104Suppose the right line geuen to be AB. It is required to deuide the line AB by an extreme and meane proportion. Deuide the line AB in the point C (by the 11. of the second) in such sort that the rectangle figure comprehended vnder the lines AB and BC may be equall vnto the square de∣scribed

[illustration]

of the line CA. And forasmuch as that which is comprehended vnder the lines AB & BC, is equall vnto the square made of the line AC, therefore as the line BA is to the line AC, so (by the 17. of the sixt) is the line AC to the lin•• CB. Wherfore the line AB is deuided by an extreme & meane proportion in the point C. which was required to be done.

The 21. Theoreme. The 31. Proposition. In rectangle triangles the figure made of the side subtending the right angle, is equal vnto the figures made of the sides cō∣prehending the right angle, so that the sayd thr••e figures b••

b•• like and in like sort described.

SVppose that there be a triangle ABC, whose angle BAC let be a right angle. Thē I say that the figure which is described of the lyne BC is equall vnto the two figures which are described of the lines BA & AC, the said thre figures being like the one to the other, and in like sort described. From the point A (by the 12. of the first) let there be drawne vn∣to the line BC a perpendiculer line AD.* 1.105 Now forasmuch as in y^e rectangle trian∣gle ABC is drawen from the right angle A vnto the base BC a perpendicular line AD,* 1.106 therfore the triangles ABD, and ADC set vpon the perpendiculer line, are like vnto the whole triangle ABC, and also like the one to the other (by the 8. of the sixt). And forasmuch as the triangle ABC is like vnto the triangle ABD, therfore as the line CB is to the line BA, so is the line AB to the lyne BD. Now for that there are three right lines proportional, therfore (by the 2. correllary of the 20. of the sixth) as the first is to the third, so is the figure made of the first, to the figure made

[illustration]

of the second, the said figures being like and in like sorte de∣scribed. Wherefore as the lyne BC is to the line BD, so is the figure made of the line BC to the figure made of the line BA, they beyng like and in lyke sort described. And by y^e same reason as the line BC is to the lyne CD, so is the figure made of the line BC, to the figure made of the line CA, they be∣ing like and in like sort descri∣bed. Wherfore as the line BC is to the lines BD and DC, so is the figure made of the line BC to the figures made of the lines BA and AC, they being like and in like sort described. But the line BC is equall vnto the lines BD and DC, wherfore the figure made of the lyne BC is equall vnto the figures made of the lines BA and AC, they being like and in like sort described. Wherfore in rectan∣gle triangles the figure made of the side subtending the right angle is equall vnto the figures made of the sides comprehending the right angle, so that the said thr•• figures be like and in like sort described: which was required to be proued.

An other way.

Forasmuch as (by the first correllary of the 20. of the sixth) like rectiline fi∣gures are in double proportion to that that the sides of like proportion are, ther∣fore

the rectiline figure made of the line BC is vnto t••e rectiline figure ••ade of the line BA in double proportion to that that the line CB is to the line BA, and (by the same) the square also made of the line BC is vnto the square made of the line BA in double proportion

[illustration]

to that that the line CB is vn∣to the line BA. Wherfore also as the rectiline figure made of the line CB is to the rectiline figure made of the line BA, so is the square made of the lyne CB to the square made of the line BA. And by the same rea¦son also, as the rectiline figure made of the line BC is to the rectiline figure made of y^e line CA, so is the square made of the line BC to the square made of the line CA. Wherfore also as the rectiline figure made of the line BC is to the rectiline figures made of the lines BA and AC, so is the square made of the line BC to the squares made of the lines BA and AC. But the square made of y^e line BC is equall vnto the squares made of the lines BA and AC (by the 47. of the first) Wherefore also the rectiline figure made of the line BC is equall vnto the rectiline figures made of the lines BA and AC, the said three figures beyng lyke and in like sort described.

The conuerse of this Proposition after Campane.

* 1.107If the figure described of one of the sides of a triangle be equall to the figures which are described of the two other sides, the sayd figures being like and in like sort described, the tri∣angle shall be a rectangle triangle.

Suppose that ABC be a triangle, and let the fi∣gure

[illustration]

described of the side BC be equall to the two figures described of the sides AB and AC, the said figures being like, and in like sort descri∣bed. Then I say that the angle A is a right angle. Let the angle CAD be a right angle, and put the line AD equall to the line AB, and drawe a lyne from D to C. Now then by this 31. proposition, the figure made of the line CD is equall to the two figures made of the lines AC and AD, the said figures being like and in like sort described. Wherefore also it is equall vnto the figure made of the line BC, which is by supposition e∣qual to the two figures made of the lines AC and AD (for the line AD is put equall to the line AB) wherfore the line DC is ••quall to the line BC. Wherfore (by the 8. of the first) the angle BAC is a right angle, which was required to be proued.

The 22. Theoreme. The 32. Proposition. If two triangles be set together at one angle, hauing two sides of the one proportionall to two sides of the other, so that their sides of like proportion be also parallels: then the other sides remayning of those triangles shall be in one right line.

SVppose the two triangles to be ABC, and DCE, and let two of their sides AC & DC make an angle ACD, and let the said triangles haue two sides of the one, namely, BA and AC proportionall to two sides of the other,* 1.108 namely, to DC and DE, so y^t as AB is to AC, so let DC be to DE. And let AB be a parallell vnto DC, and AC a parallell vnto DE. Then I say, that the lines BC and CE are in one right line. For forasmuch as the line AB is a parallell vnto the line DC, and

[illustration]

vpon thē lighteth a right line AC:

[illustration]

therefore (by the 29. of the first) the alternate angles BAC and ACD are equall the one to the other. And by the same reason the angle CDE is equall vnto y^e same angle ACD. Wherefore the angle BAC is equall vnto the angle CDE. And foras∣much as there are two triangles ABC and DCE, hauing the angle A of the one equall to the angle D of the other, and the sides about the equall angles are (by supposition) proportionall, that is, as the line BA is to the line AC, so is the line CD to the line DE, ther∣fore the triangle ABC is (by the 6. of the sixt) equiangle vnto the triangl•• DCE. Wherefore the angle ABC is equall vnto the angle DCE. And it is proued, that the angle ACD is equall vnto the angle BAC. Wherefore the whole angle ACE is equall vnto the two angles ABC and BAC. Put the an∣gle ACB common to them both. Wherefore the angles ACE and ACB are e∣quall vnto the angles CAB, ACB, & CBA. But the angles CAB, ACB, and CBA, are (by the 32. of the first) equall vnto two right angles. Wherefore also the angles ACE and ACB are equall to two right angles. Now then vn∣to the right line AC, and vnto the point in it C, are drawen two right lines BC and CE, not on one and the same side, making the side angles ACE & ACB equall to two right angles. Wherefore the lines BC and CE (by the 14. of the first) are set directly and do make one right line. If therefore two triangles be set together at one angle hauing two sides of the one proportionall to two sides of y^e other, so y^t their sides of like proportion be also parallels: then y^e sides remay∣ning of those triangles shall be in one right line: which was required to be proued.

Although Euclide doth not distinctly set forth the maner of proportion of like rectiline figures, as he did of lines in the 10. Propositiō of this Booke, and in the 3. following it, yet as Flussates noteth, is that not hard to be done by the 22. of thys Booke. ••or two like rectiline figures being geuen to finde out a third proportio∣nall•• also betwene two rectiline superficieces geuen to finde out a meane propor∣tionall (which we before taught to do by Pelitarius after the 24. Proposition of this booke): and moreouer three like rectiline figures being geuen to finde out a fourth proportionall like and in like sort described, and such kinde of proportions, are easie to be found out by the proportions of lines. As thus. If vnto two sides of like proportion we should find out a third proportionall by the 11. of this boke•• the rectiline figure described vpon that line shall be the third rectiline figure pro∣portionall with the two first figures geuen by the 22. of thys booke. And if be∣twene two sides of like proportion be taken a meane proportionall by the 13. of thys Booke: the rectiline ••igure described vpon the sayd meane shall likewise be a meane proportionall betwene the two rectiline figures geuē by the same 22. of the sixt. And so if vnto three sides ge••en be found out the fourth side proportionall (by the 12. of this booke) the rectiline ••igure described vpon the sayd fourth line shall be the fourth rectiline figure proportionall. For if the right lines be proporti∣onall, the rectiline figures described vpon them shall also be proportionall, so that the said rectiline ••igures be like & in like sort described by the said 22. of the sixt.

The 23. Theoreme. The 33. Proposition. In equal circles, the angles haue one and the selfe same pro∣portion that the circumferēces haue, wherin they cōsist, whe∣ther the angles be set at the centres or at the circumferences. And in like sort are the sectors which are described vppon the centres.

SVppose the equall circles to be ABC, and DEF, whose centres let be G and H, and let the angles set at their centres G and H, be BGC, and EHF: and let the angles set at their circumferences be BAC and EDF. Then I say that as the circumference BC is to the circumference EF so is the angle BGC to the angle EHF: and the angle BAC to the angle EDF, and moreouer the sector GBC to the sector HEF. Vnto the circumference BC (by the 38. of the third) put as many equall circumferences in order as you will, namely, CK and KL, and vnto the circumference EF put also as many e∣quall circumferences in number as you will, namely, FM, and MN. And drawe these right lines GK,* 1.109 GL, HM, and HN. Now forasmuch as the circumfe∣rences BC, CK, and KL are equall the one to the other, the angles also BGC, and CGK, and KGL are (by the 27. of the third) equall the one to the other. Therfore how multiplex the circumference BL is to the circumference BC, so multiplex is y^e angle BGL to the angle BGC. And by y^e same reason also, how multiplex the circumference NE is to the circumference EF, so multiplex is the

angle NHE to the angle EHF. Wherfore if the circumference BL be equal vnto the circumference EN, the angle BGL is equall vnto the angle EHN, and if the circumference BL be greater then the circumference EN, the angle BGL is greater then the angle NHE, and if the circumference be lesse, the angle is lesse. Now then there are foure magnitudes, namely, the two circumfe∣rences BC and EF, and the two angles that is, BGC, and EHF, and to the circumference BC and to the angle BGC, that is, to the first and third are takē equemultiplices, namely, the circumference BL, and the angle BGL, and like∣wise to the circumference EF, and to the angle EHF, that is, to the second and fourth, are taken certayne other equemultiplices, namely, the circumference EN and the angle EHN. And it is proued, that if the circumference BL exceede the circumference EN, the angle also BGL exceedeth the angle EHN. And if the circumference be equall, the angle is equall, and if the circumferēce be lesse, the angle also is lesse. Wherfore (by the 6. definition of the fifth) as the circumfe∣rence BC is to the circumference EF, so is the angle BGC to the angle EHF••

[illustration]

But as the angle BGC is to the angle EHF,* 1.110 so is the angle BAC to the angle EDF, for the angle BGC is double to the angle BAC, and the angle EHF is also double to the angle EDF (by the 20. of the third) Wherfore as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. Wherfore in equall circles the angles are in one and the selfe same proportion that their circumferences are, whether the angles be set at the centres, or at the circumferences: which was re∣quired to be proued.

I say moreouer that as the circumference BC is to the circumference EF,* 1.111 so is the sector GBC to the sector HEF. Draw these lines BC and CK. And in the circumferences BC and CK take pointes at all aduentures, namely, P and O. And draw lines from B to P, and from P to C, from C to O, and from O to K. And forasmuch as (by the 15. definition of the first) the two lines BG and GC are equall vnto the two lines CG and GK, and they also comprehend equall an∣gles,

[illustration]

therfore (by the 4. of the first) the base BC is equall vnto the base CK, & the triangle GBC is equall vnto the triangle GCK. And seing that the circum¦ference BC is equall vnto the circumference CK, therefore the circumference remayning of the whole circle ABC, namely, the circumference BLAKOC, is equall vnto the circum••erence remayning of the selfe same circle ABC, name∣ly, to the circumference CPBLAK. Wherfore the angle BPC is equall vnto the angle COK (by the 27. of the third) Wherfore (by the 10. definition of the third) the segment BPC is like vnto the segment COK, and they are set vpon equall right lines BC and KC. But like segmentes of circles which consist vpon equall right lines are also equall the one to the other (by the 24. of the third). Wherfore the segment BPC is equall vnto the segment COK. And the trian∣gle GBC is equall vnto the triangle GCK. Wherfore the sector GBC is equall vnto the sector GCK. And by the same reason also, the sector GKL is e∣quall vnto either of the sectors GBC and GCK. Wherfore y^e three sectors GBC and GCK, and GKL, are equall y^e one to y^e other. And by the same reason also the sectors HEF and HFM, and HMN are equall the one to the other. Wherefore how multiplex the circumference BL is to the circumference BC so multiplex is the sector GLB to the sector GBC. And by the same reason how multiplex the circumference NE is to the circumference EF, so multiplex is y^e sector HEN to the sector HEF. If therfore the circumference BL be equall vnto y^e circumference EN, y^e sector also BGL is equall vnto y^e sector EHN. And if the circumferēce BL excede y^e circumference EN, y^e sector also BGL excedeth the sector EHN. And if the circumference be lesse, the sector also is lesse. Now thē there are foure magnitudes, namely, the two circumferences BC and EF, and the two sectors GBC & HEF, and to the circumference BC, & to the sector GBC, namely, to the first and the third, are taken equemultiplices, that is, the circumference BL, and the sector GBL, and likewise to the circum∣ference EF, and to the sector HEF, namely, to the second and fourth, are ta∣ken certayne other equimultiplices, namely, the circumference EN and the sec∣tor

HEN. And it is proued that if the circumference BL excede the circumfe∣rence EN, the sector also BGL excedeth the sector EHN. And if the circumference be equall, the segment also is equall, and if the circumference be lesse, the segment also is lesse. Wherfore (by the conuersion of the sixt definition of the fifth) as the circumference BC is to the circumference EF, so is y^e sector GBC vnto the sector HEF: which is all that was required to be proued.

Corollary. And hereby it is manifest, that as the sector is to the sector, so is angle to angle by the 11. of the fifth.

Flussates here addeth fiue Propositions wherof one is a Probleme hauing three Corollaryes following of it, and the rest are Theoremes : which for that they are both witty, & also serue to great vse, as we shall afterward see, I thought not good to omitte, but haue here placed them: but onely that I haue not put them to fol∣lowe in order with the Propositions of Euclide as he hath done.

¶The first Proposition added by Flussates.

To describe two rectiline figures equall and like vnto a rectiline figure geuen and in like sort situate, which shall haue also a proportion geuen.

Suppose that the rectiline figure geuen be ABH.* 1.112 And let the proportion geuen be the proportion of the lines GC and CD. And (by the 10. of this booke) deuide the line AB like vnto the line GD in the poynt E (so that as the line GC is to the line CD, so let the line AE be to the line EB). And vpon the line AB describe a semicircle AFB. And from the poynt E erect (by the 11. of the first) vnto the line AB a perpendicular line EF cutting the circumference in the poynt F. And draw these lines AF and FB. And vpon either of these lines describe rectiline figures like vnto the rectiline figure AHB and in like sort situate (by the 18. of the sixt): which let be AKF, & FIB. Then I say, that the rectiline figures AKF, and FIB,

[illustration]

haue the proportion geuē (namely, the propor∣tion of the line GC to the line CD) and are e∣quall to the rectiline figure geuen ABH vnto which they are described like and in like sort si∣tuate. For forasmuch as AFB is a semicircle, therefore the angle AFB is a right angle (by the 31. of the third) and FE is a perpendicular line.* 1.113 Wherefore (by the 8. of this booke) the triangles AFE and FBE are like both to the whole triangle AFB and also the one to the other. Wherefore (by the 4. of this booke) as the line AF is to the line FB, so is the line AE to the line EF, and the line EF to the line EB, which are sides cōtayning equall angles. Wher∣fore (by the 22. of this booke) as the rectiline fi∣gure described of the line AF is to the rectiline figure described of the line FB, so is the recti∣line figure described of the line AE to the recti∣line figure described of the line EF, the sayd rectiline figures being like and in like sort

••ituate. But as the rectiline figure described of the line AE being the ••irst, is to the rectiline ••igure described of the line EF being the second, so is the line AE the first, 10. the line ••B the third (by the 2. Corollary of the 20. of thys booke). Wherfore the recti∣line figure described of the line AF is to the rectiline figure described of the line FB, as the line A•• is to the line EB. But the line AE is to, the line EB (by construction) as the line GC is to the line CD. Wherefore (by the 11. of the fift) as the line GC is to the line CD, so is the r••••tiline ••igure described of the line AF to the rectiline ••igure described 〈◊〉〈◊〉 the line ••B, the sayd rectiline figures being like and in like sort described. But the 〈…〉〈…〉 described o•• the lines AF and FB, are equall to the rectiline ••igure d••••••••••bed o•• the line AB, vnto which they are (by construction) described lyke and in like sort situate. Wherefore there are described two rectiline figures AKF and FIB equ••ll and like vnto the rectiline figure geuen ABH and in like sort situate, and they ha••e also the one to the other the proportion geuen, namely, the proportion of the line GC to the line CD: which was required to be done.

¶The first Corollary.

To resolue a rectiline figure geue into two like rectiline ••igures which shall hau•• also a proporti•• ge∣••ē. For i•• there be put three right lines in the proportiō geuē, and if the line AB be cu•• in the same proportion that the first line is to the third,* 1.114 the rectiline ••igures described of the line•• A•• and FB (which figures haue the same proportion that the lines AE and EB haue) shall be in double proportion to that which the lines AF and FB are (by the ••irs•• ••orollary o•• the 20. o•• this booke). Wherefore the right lines AF and FB are the o••e to the other in the same proportion that the first of the three lines put is to the 〈◊〉〈◊〉. ••or t••e 〈◊〉〈◊〉 line to the third, namely, the line AE to the line EB is in dou∣ble propo••tion that it is to the second, by the 10. de••inition of the fi••t.

¶ The second Corollary.

Hereby may we learne, how from a rectiline ••igure geuen to take away a part appointed, lea••ing, the rest of the rectiline ••igure like vnto the whole.* 1.115 For if frō the right line AB be cut of a part appoynted, namely, EB (by the 9. of this booke) as the line AE is to the line EB, so is the rectiline ••igure described of the line AF to the rectiline figure described of the line FB (the sayd ••igures being supposed to be like both the one to the other and also to the rectiline ••igure described of the line AB, and being also in like sort situate). Wherfore taking away ••rom the rectiline ••igure described of the line AB, the rectiline figure de∣scribed of the line FB, the residue, namely, the rectiline figure described of the line AF shall be both like vnto the whole rectiline ••igure geuen described of the line AB, and in like sort situate.

¶ The third Corollary.

To compose two like rectiline ••igures into one rectiline figure like and equall to the same figures. Let their sides of like proportiō be set so that they make a right angle,* 1.116 as the lines AF and FB are. And vpō the line subtending the said angle, namely, the line AB, describe a rectiline ••igure like vnto the rectiline figures geuen and in like sort situate (by the 18. of this booke) and the same shall be equall to the two rectiline figures geuen (by the 31. of this booke).

¶ The second Proposition.

If two right lines cut the one the other obtuseangled wise, and from the endes of the lines which ••ut the one the other be drawen perpendicular lines to either line: the lines which are betwene the endes and the perpendicular lines are cut reciprokally.

Suppose that there be two right lines AB and GD cutting the one the other in the point E, and making an obtuse angle in the section E. And from the endes of the lines, namely, A and G, let there be drawen to either line perpendicular lines, namely, from the point A to the line GD, which let be AD, and from the point G to the right line AB: which let be GB. Then I say, that the

[illustration]

right lines AB and GD do, betwene the end A and the perpendicular B, and the end G and the perpendicular D, cut the one the other reciprokally in the point E: so that as the line AE is to the line ED, so is the line GE to the line EB.* 1.117 For foras∣much as the angles ADE and GBE, are right angles, therfore they are equall. But the angles AED and GEB are also e∣quall (by the 15. of the first). Wherefore the angles remayning, namely, EAD, & EGB, are equall (by the Corollary of the 32. of the first). Wherefore the triangles AED and GEH, are equiangle. Wherfore the sides about the equall angles shall be proportionall (by the 4. of the sixt). Wherfore as the line AE is to the line ED, so is the line GE to the line EB. If therefore two right lines cut the one the other obtuseangled wife. &c: which was required to be proued.

¶ The third Proposition.

If two right lines make an acute angle, and from their endes be drawen to ech line perpendicular lines cutting them: the two right lines geuen shall be reciprokally pro∣portionall as the segmentes which are about the angle.

Suppose that there be two right lines AB and GB, making an acute angle ABG. And from the poyntes A and G let there be drawen vnto the lines AB and GB per∣pendicular lines AC and GE, cutting the lines AB and GB in the poyntes E and ••. Then I say, that the lines, namely AB to GB, are reciprokally proportionall, as the seg∣mentes, namely, CB to EB which are about the acute

[illustration]

angle B.* 1.118 For forasmuch as th•• right angles ACB and GER are equall, and the angle•• ABG is common to the triangles ABC, and GBE•• therefore the angles re∣mayning BAC and EGB are equall (by the Corolla∣ry of the 32. of the first). Wherfore the triangles ABC and GBE are equiangle. Wherefore the side•• about the equall angles are proportionall (by the 4. of the sixe)•• so that, as the line AB is to the line FC, so is the line GB to the line BE. Wherefore alternately as the line AB is to the line GB so is the line CB to the line BE. If therefore two right lines mak•• a•• ••c••te angle•• &c•• which was required to be proued.

•• The fourth Proposition••

If in a circle be drawen two right lines cutting the one the other, the sections of the one to the sections of the other shall be reciprokally proportionall.

In the circle, AGB let these two right lines 〈…〉〈…〉 one the other in the poynt E.* 1.119 Th•••• I say, that reciprokally 〈◊〉〈◊〉 ••h•• line AE is to the line ED, so is the line GE to the line EB. For forasmuch as (by the 35. of the third) the rectangle fi∣gure

contayned vnder the lines AE and

[illustration]

EB is equall to the rectangle figure contay∣ned vnder the lines GE and ED, but in e∣quall rectangle parallelogrammes the sides about the equall angles are reciprokall (by the 14. of the sixt). Therefore the line AE is to the line ED reciprokally as the line GE is to the line EB (by the second defi∣nition of the sixt). If therefore in a circle be drawen two right lines. &c: which was required to be proued.

¶The fift Proposition.

If from a poynt geuen be drawen in a plaine super••icies two right lines to the concaue circumference of a circle: they shall be reci∣prokally proportionall with their partes takē without the circle. And moreouer a right line drawen from the sayd poynt & touching the circle, shall be a meane proportionall betwene the whole line and the vtter segment.

Suppose that there be a circle ABD, and without it

[illustration]

take a certayne poynt, namely, G. And from the point G drawe vnto the concaue circumference two right lines GB and GD, cutting the circle in the poyntes C and E. And let the line GA touch the circle in the point A. Thē I say, that the lines, namely, GB to GD are reciprokally as their parts taken without the circle, namely, as GC to GE.* 1.120 For forasmuch as (by the Corollary of the 36. of the third) the rectangle figure contayned vnder the lines GB and GE is equall to the rectangle figure contayned vn∣der the lines GD and GC, therefore (by the 14. of the sixt) reciprokally as the line GB is to the line GD, so is the line GC to the line GE, for they are sides contayning equall angles. I say moreouer, that betwene the lines GB and GE, or betwene the lines GD and GC the touch line GA is a meane proportionall.* 1.121 For forasmuch as the re∣ctangle figure comprehended vnder the lines GB and GE is equall to the square made of the line AG (by the 36. of the third) it followeth that the touch line GA is a meane propor∣tionall betwene the extremes GB and GE (by the second part of the 17. of the sixt) for that by that Proposition the lines GB, GA, and GE are proportionall. And by the same reason may it be proued that the line GA is a meane proportionall betwene the lines GD and GC, and so of all others. If therefore from a poynt geuen•• &c•• which was required to be demonstrated.

The end of the sixth booke of Euclides Elementes.

Notes

• * 1.1

  The argument of this sixth booke.
• * 1.2

  This booke necessary for the vse of in∣strumentes of Geometry.
• * 1.3

  The first de∣finition.
• * 1.4

  The second de••inition.
• * 1.5

  Reciprocall figures called mutuall fi∣gures.
• * 1.6

  The third de∣finition.
• * 1.7

  The fourth definition.
• * 1.8

  The fifth de∣finition.
• * 1.9

  By the name of quantities is vnderstan∣ded the deno∣minations of proportions.
• * 1.10

  Example of this definitiō.
• * 1.11

  Example in numbers.
• * 1.12

  2. 3. 15. 18.
• * 1.13

  An other example.
• * 1.14

  Of substracti∣on of propor∣tion.
• * 1.15

  The sixth de∣finition.
• * 1.16

  Construction.
• * 1.17

  Demonstra∣tion of the first part.
• * 1.18

  Demonst••a∣tion of the second part.
• * 1.19

  A Corollary added by Flussates.
• * 1.20

  The first part of this Theo∣reme.
• * 1.21

  Demonstra∣tion of the second part.
• * 1.22

  A Corollary added by Flussates.
• * 1.23

  Construction.
• * 1.24

  Demonstra∣tion of the first part.
• * 1.25

  Demonstra∣ti••n of the second part, which is the conuerse of the first.
• * 1.26

  Construction.
• * 1.27

  Demonstra∣tion.
• * 1.28

  This is the conuerse of the former proposition.
• * 1.29

  Construction
• * 1.30

  Demonstra∣tion.
• * 1.31

  Constructio••.
• * 1.32

  The first part of this proposition.
• * 1.33

  Demonstra∣tion leading to an impossibi∣litie.
• * 1.34

  The second part of this proposition.
• * 1.35

  Construction.
• * 1.36

  Demonstra∣tion.
• * 1.37

  Construction.
• * 1.38

  Demonstra∣tion.
• * 1.39

  Construction.
• * 1.40

  Demonstra∣tion.
• * 1.41

  A Corollary out of Flussates.
• * 1.42

  By this and the former propo••ition may a right line be deui∣ded into what partes soeuer you will.
• * 1.43

  Construction.
• * 1.44

  Demonstra∣tion.
• * 1.45

  An other way after Pelita¦rius.
• * 1.46

  An ot••e•• way after Pelitarius.
• * 1.47

  Construction.
• * 1.48

  Demonstra∣tion.
• * 1.49

  An other way after Campane.
• * 1.50

  Construction.
• * 1.51

  Demonstra∣tio••.
• * 1.52

  A proposition added by Pe∣litarius.
• * 1.53

  The first part of this propo∣sition.
• * 1.54

  Demonstra∣tion of the of the same.
• * 1.55

  The second part which is the conuerse of the first.
• * 1.56

  The first par•• of this propo∣sition.
• * 1.57

  Demonstra∣tion of the same.
• * 1.58

  The second part which is the conuerse of the first.
• * 1.59

  Demonstra∣tion of the first part.
• * 1.60

  The second part which is the conuerse of the first.
• * 1.61

  The first part of th•••• Theoreme.
• * 1.62

  The second part which is the conuerse of the first.
• * 1.63

  A Co••ollary.
• * 1.64

  Description of the rectiline figure r••qui∣red.
• * 1.65

  Demonstra∣tion.
• * 1.66

  Demonstra∣tion.
• * 1.67

  A Corollary.
• * 1.68

  The first par•• of this Theoreme.
• * 1.69

  The second part demon∣strated.
• * 1.70

  The third part.
• * 1.71

  The first Co∣rollary.
• * 1.72

  The second Corollary.
• * 1.73

  Demonstra∣tion.
• * 1.74

  The first part of this proposition.
• * 1.75

  The second part which is the conuerse of the first.
• * 1.76

  Note that this is proued in the assumpt following.
• * 1.77

  An Assumpt.
• * 1.78

  An other demō∣stration of the second part af∣ter Flussates.
• * 1.79

  Demonstra∣tion.
• * 1.80

  An other demonstration after flussates.
• * 1.81

  Demonstration of this propositiō wherein is first proued that the parallegramme EG is like to the whole parallelo∣grāme ABCD.
• * 1.82

  That the paral∣lelogrāme KH is like to the whole parallelo∣gramme AB∣CD.
• * 1.83

  That the paral∣lelogrammes EG and KH are like the one to the other.
• * 1.84

  An other Demonstra∣tion after Flussates.
• * 1.85

  An addition of Pelitarius.
• * 1.86

  Another ad∣dition of Pelitarius.
• * 1.87

  Construction.
• * 1.88

  Demonstra∣tion.
• * 1.89

  Demonstra∣tion.
• * 1.90

  By the di∣metiēt is vnderstand here the dime∣tient which is ••rawen from the angle which is com∣mon to them both to the op∣posite angle.

  Demonstra∣tion leading to an absurditie.
• * 1.91

  An other way after Flussates.
• * 1.92

  In this propositiō are two cases, in the first the pa∣rallelogramme compared to the parallelogrāme described of the halfe line is de∣scribed vpon a line greater thē the halfe line: In the second vpō a line lesse.

  The first case where the pa∣rellelogramme compared name¦ly AF is descri∣bed vpon the line AK which is greater then the halfe line AC.
• * 1.93

  Demonstra∣tion of this case.
• * 1.94

  The second case where the parallelo∣gramme compared namely AE is described vpon the line AD which is lesse then the line AC, Demonstrati∣on of the se∣cond case.
• * 1.95

  Construction.
• * 1.96

  Two cases in this Proposi∣tion.
• * 1.97

  The first case.
• * 1.98

  The second case.
• * 1.99

  A Corollary ad∣ded by Flussa∣tes, and is put of Theon as an assumpt be••ore the 17. proposi∣tion of the tēth booke: which ••or that it follo∣weth of this proposition I thought it not amisse here to place.
• * 1.100

  Construction.
• * 1.101

  Demonstra∣tion.
• * 1.102

  Construction.
• * 1.103

  Demo••stra∣••ion.
• * 1.104

  An other way.
• * 1.105

  Construction.
• * 1.106

  Demonstra∣tion.
• * 1.107

  The conuerse of the former proposition.
• * 1.108

  Demonstra∣tion.
• * 1.109

  That the an∣gles at th•• cē∣ter are in pro∣portiō the one to the other, as the circumfe∣rences wheron they are.
• * 1.110

  That the an∣gles at the cir∣cumferences are so also.
• * 1.111

  That the sec∣tors are so also.
• * 1.112

  Construction of the Pro∣bleme.
• * 1.113

  Demonstar∣tion of the same.
• * 1.114

  The first Co∣rollary.
• * 1.115

  The second Corollary.
• * 1.116

  The third Corollary.
• * 1.117

  Demonstra∣tion of this proposition.
• * 1.118

  Demonstra∣tion of this propositions
• * 1.119

  Demonstrati∣on of this pro∣position.
• * 1.120

  Demonstrati∣on of the first part of this proposition.
• * 1.121

  Demonstra∣tion of the se∣cond part.