The elements of Euclid, explained and demonstrated in a new and most easie method with the uses of each proposition in all the parts of the mathematicks / by Claude Francois Milliet D'Chales, a Jesuit ; done out of French, corrected and augmented, and illustrated with nine copper plates, and the effigies of Euclid, by Reeve Williams ...

PROPOSITION I. THEOREM.

PArallelograms and Triangles of equal height have the same Reason as their Bases.

* 1.1 Let there be proposed the Triangles ACG, DEM, of equal height, in such sort that they may be placed between the same Parallels AD, GM: I say that there shall be the same Reason of the Base GC to the Base EM, as of the Triangle AGC, to the Triangle DEM. Let the Base EM be divided into as many equal parts as you please, and let there be drawn to each Division the Lines DF, DH, &c. Let also the Line GC be divided into parts equal to those of EM, and let be drawn Lines to each division from the top A: All those little Triangles which are made in the Two great Triangles, are be∣tween the same Parallels, and they have equal Bases; they are thence equal (by the 28th.)

Demonstration. The Base GC, con∣tains as many Aliquot parts of the Line EM, as could be found parts equal to

EF: Now as many times as there are in the Base GC parts equal to EF; so many times the Triangle AGC con∣taineth the little Triangles equal to those which are in the Triangle DEM; which being equal among themselves, are its Aliquot parts: thence as many times as the Base GC containeth Ali∣quot parts of EM, so many times the Triangle AGC containeth Aliquot parts of the Triangle DEM; which will happen in all manner of Divisions. There is thence the same Reason of the Base GC to the Base EM, as of the Triangle AGC to the Triangle DEM.

Coroll. Parallelograms drawn on the same Bases, and that are between the same Parallel Lines, are double to the Triangle (by the 41st.) they are thence in the same Reason, as Triangles, that is to say, in the same Reason as their Bases.

USE.

* 1.2 THis Proposition is not only necessary to Demonstrate those which follow, but we may make use thereof in Dividing of Land. Let there be proposed a Tra∣pezium

ABCD, which hath its sides AD, BC, parallel, and admit one would cut of a third part: Let CE be made equal to AD; and BG the third part of BE. Draw AG. I say that the Tri∣angle ABG is the Third of the Trape∣zium ABCD.

Demonstration The Triangles ADF, FCE, are equiangled, because of the Pa∣rallel AD, CE; and they have their Sides AD, CE, equal: They are thence equal (by the 26th. of the 1st.) and by consequence the Triangle ABE is equal to the Trapezium. Now the Triangle ABG is the third part of the Triangle ABE by the preceding Proposition; thence the Triangle ABG is the third part of the Trapezium ABCD.

PROPOSITION II. THEOREM.

A Line being drawn in a Triangle Parallel to its Base, divideth its sides proportionally, and if a Line divi∣deth proportionally the sides of a Triangle, it shall be parallel to its Base.

If in the Triangle ABC, the Line

DE, is Parallel to the Base BC; the Sides AB, AC, shall be divided pro∣portionally; that is to say, that there shall be the same Reason of AD to DB, as of AE to EC. Draw the Lines DE, BE. The Triangles DBE, DEC, which have the same Base DE, and are between the same Parallels DE, BC; are equal (by the 37th. of the 1st.)

Demonstration. The Triangles ADE, DBE, have the same point E for their vertical, if we take AD, DB, for their Bases; and if one should draw through the point E, a Parallel to AB, they would be both between the same Parallels; they shall have thence the same Reason as their Bases (by the 1st.) that is to say, that there is the same Reason of AD to DB, as of the Triangle ADE to the Triangle DBE, or to its equal CED. Now there is the same Reason of the Triangle ADE to the Triangle CED, as of the Base AE to EC. There is therefore the same Reason of AD to DB, as of AE to EC.

And if there be the same Reason of AE to EC, as of AD to DB: I say that the Lines DE, BC, would then be Parallels.

Demonstration. There is the same Reason of AD to DB, as of the Tri∣angle ADE to the Triangle DBE (by the 1st.) there is also the same Reason of AE to EC, as of the Triangle ADE to the Triangle DEC; consequently there is the same Reason of the Triangle ADE to the Triangle BDE, as of the same Triangle ADE to the Triangle CED. So then, (by the 7th. of the 5th.) the Triangles BDE, CED, are equal: And (by the 39th. of the 1st.) they are between the same Parallels.

USE

THis Proposition is absolutely necessary, in the following Propositions, one may make use thereof in Measuring as in the following figure: If it were required to measure the height BE; having the length of the staff DA, there is the same Reason of CD to DA, as of BC to BE.

PROPOSITION III. THEOREM.

THat Line which divideth the Angle of a Triangle, into two equal parts, divideth its Base in two parts, which are in the same Reason to each other as are their Sides. And if that Line divideth the Base into parts proportional to the Sides, it shall divide the Angle into Two equally.

If the Line AD divideth the Angle BAC into Two equal parts; there shall be the same Reason of AB to AC, as of BD to DC. Continue the Side CA, and make AE equal to AB; then draw the Line EB.

Demonstration. The exterior Angle CAB is equal to the Two interior Angles AEB, ABE; which being equal (by the 5th. of the 1st.) seeing the Sides AE, AB, are equal; the Angle BAD, the half of BAC, shall be equal to one of them; that is to say to the Angle ABE. Thence (by the 27th. of the 1st.) the Lines AD, EB, are parallel; and (by the 2d.) there is the same Reason of EA, or AB to AC, as of BD to DC.

Secondly. If there be the same Reason of AB to AC, as of BD to DC, the Angle BAC shall be divided into Two equally.

Demonstra. There is the same reason of AB or AE to AC, as of BD to DC: thence the Lines EB, AD, are parallel; and (by the 29th. of the 1st.) the Alternate Angles EBA, BAD, the internal BEA, and the external DAC, shall be equal; and the Angles EBA, AEB, being equal; the Angles BAD, DAC, shall be so likewise. Wherefore the Angle BAC hath been divided equally.

USE.

WE make use of this Proposition to attain to the Proportion of the sides.

PROPOSITION IV. THEOREM.

EQuiangular Triangles have their Sides Proportional.

If the Triangles ABC, DCE, are equiangular; that is to say, that the

Angles ABC, DCE; BAC, CDE, be equal: There will be the same Rea∣son of BA to BC, as of CD to CE. In like manner the reason of BA to AC, shall be the same with that of CD to DE. Joyn the Triangles after such a manner, that their Bases BC, CE, be on the same Line; and continue the sides ED, BA: seeing the Angles ACB, DEC, are equal; the Lines AC, EF, are parallel; and so CD, BF, (by the 29th. of the 1st.) and AF, DC, shall be a parallelogram.

Demonstration. In the Triangle BFE, AC, is parallel to the Base FE, thence (by the 2d.) there shall be the same reason of BA to AF, or CD, as of BC to CE; (and by exchange) there shall be the same reason of AB to BC, as of DC to CE. In like manner in the same Triangle, CD being parallel to the Base BF; there shall be the same Reason of FD, or AC to DE, as of BC to GE (by the 2d.) and by ex∣change, there shall be the same reason of AC to BC, as of DE to CE.

USE.

THis Proposition is of a great extent, and may pass for a universal Princi∣ple in all sorts of Measuring. For in the first place the ordinary practice in measu∣ring inaccessible Lines, by making a little Triangle like unto that which is made or imagined to be made on the ground, is founded on this Proposition, as also the greatest part of those Instruments, on which are made Triangles like unto those that we would measure, as the Geometrical Square, Sinical Quadrant, Jacobs Staff, and others. Moreover we could not take the plane of a place, but by this Proposi∣tion: wherefore to explain its uses, we should be forced to bring in the first Book of practical Geometry.

PROPOSITION V. THEOREM.

TRiangles whose sides are proportional, are equianguler.

If the Triangles ABC, DEF, have their sides proportional, that is to say, if there be the same reason of AB to

BC, as of DE to EF; as also if there be the same reason of AB to AC, as of DE to DF, the Angles ABC, DEF, A and D; C and F shall be equal. Make the Angle FEG equal to the Angle B; and EFG equal to the Angle C.

Demonstration. The Triangles ABC, EFG, have two Angles equal; they are thence equiangled (by the Cor. of the 32d. of the 1st.) and (by the 4th.) there is the same reason of DE to EF, as of EG to EF. Now it is supposed that there is the same reason of DE to EF, as of EG to EF. Thence (by the 7th. of the 5th.) DE, EG, are equal. In like manner DF, FG, are also equal, and (by the 8th. of the 1st.) the Triangles DEF, GEF, are equiangular. Now the Angle GEF was made equal to the Angle B: thence DEF is equal to the Angle B; and the Angle DFE, to the Angle C. So that the Triangles ABC, DEF, are equiangular.

PROPOSITION VI. THEOREM.

TRiangles which have their sides propor∣tional, which include an equal Angle, are equiangular.

If the Angles B and E of the Tri∣angles ABC, DEF, being equal, there be the same reason of AB to BC, as of DE to EF; the Triangles ABC, DEF, shall be equiangular. Make the Angle FEG, equal to the Angle B, and the Angle EFG, equal to the Angle C.

Demonstra. The Triangles ABC, EGF, are equiangular (by the Cor. of the 32d. of the 1st.) there is thence the same reason of AB, to BC, as of EG to EF, (by the 4th.) Now as AB to BC, so is DE to EF; there is then the same reason of DE to EF, as of GE to EF. So then (by the 7th. of the 5th) DE, EG, are equal; and the Triangles DEF, GEF, which have their Angles DEF, GEF, each of them equal to the Angle B, and the Sides DE, EG, equal, with the Side EF common; they shall be equal in every respect (by the 4th. of the 1st.) they are thence equiangular; and the Triangle EGF being equal to the Triangle ABC, the Triangles ABC, DEF, are equiangular.

The Seventh Proposition is unnecessary.

PROPOSITION VIII. THEOREM.

A Perpendicular being drawn from the Right Angle of a Right Angled Tri∣angle to the opposite side, divideth the same into Two Triangles which are a like thereto.

If from the Right Angle ABC be drawn a perpendicular BD to the oppo∣site side AC, it will divide the Right Angled Triangle ABC into Two Trian∣gles ADB, BDC, which shall be like, or equiangular to the Triangle ABC.

Demonstration. The Triangles ABC, ADB, have the same Angle A; the Angle ADB, ABC, are right: they are thence equiangular (by the Cor. 2. of the 32d. of the 1st.) In like manner the Triangles BDC, ABC, have the Angle C common; and the Angles ABC, BDC, being right, they are also equal. Thence the Triangles ABC, DBC, are like.

USE.

WE measure inaccessable distances by a Square, according to this

Proposition. For example, if we would measure the distance DC, having drawn the perpendicular DB, and having put a Square at the Point B, in such manner, that by looking over one of its Sides BC, I see the Point C, and over its other Side I see the Point A; it is evident that there will be the same reason of AD to DB, as of DB to DC. So that multiplying DB by its self, and dividing that product by AD, the Quotient shall be DC.

PROPOSITION IX. PROBLEM.

To cut off from a Line any part required.

LEt there be proposed the Line AB, from which it is required to cut off three Fifths. Make the Angle ECD at discretion, take in one of those Lines CD, five equal parts; and let CF be three of the same, and CE be equal to AB. Then draw the Line DE; after which draw FG parallel to DE; the Line CG will contain three Fifths of CE or AB.

Demonstration. In the Triangle ECD, FG being parallel to the Base DE, there will be the same Reason of CF to FD, as of CG to GE (by the second.) and by composition (by the 18th. of the 5th.) there shall be the same Reason of CG to CE, as of CF to CD. Now CF contains three fifths of CD: wherefore CG shall contain three fifths of CE, or AB.

PROPOSITION X. PROBLEM.

TO divide a Line after the same manner as another Line is divided.

If one would divide the Line AB after the same manner as the Line AC is divided: Joyn those Lines, making an Angle at pleasure, as CAB: Draw the Line BC, and the parallels EO, FV, and the Line AB shall be divided after the same manner as AC.

Demonstration. Seeing that in the Triangle BAC, the Line HX hath been drawn parallel to the Base BC; it will divide the Sides AB, AC, pro∣portionally (by the second) it is the same with all the other parallels.

To do the same with more facility, one may draw BD parallel to AC, and put off the same Divisions of AC on BD; then draw the Lines from the one to the other.

PROPOSITION XI. THEOREM.

TO find a third Proportional to Two given Lines.

It is required to find a third propor∣tional to the Lines AB, BC, that is to say, that there may be the same reason of AB to BC, as of BC to the Line re∣quired. Make at discretion the Angle EAC; put off one after the other, the Lines AB, BC; and let AD be equal to BC: Draw the Lines BD, and its parallel CE. The Line DE shall be that which you require.

Demonstration. In the Triangle EAC, the Line DB is parallel to the Base CE: There is thence (by the 2d.) the same reason of AB to BC, as of AD, or BC to DE.

PROPOSITION XII. PROBLEM.

TO find a fourth Proportional to three Lines given.

Let there be proposed three Lines AB, BC, DE, to which must be found a fourth proportional, make an Angle as FAC, at discretion; take on AC the Lines AB, BC; and on AF, the Line AD equal to DE: then draw DB, and its parallel FC. I say that DF is the Line you seek for; that is to say, that there is the same Reason of AB to BC, as of DE, or AD to DF.

Demonstration. In the Triangle FAC, the Line DB is parallel to the Base FC; there is thence the same reason of AB to BC, as of AD to DF (by the 2d.)

USE.

THe use of the Compass of Proportion (or Sector) is established on these Propositions: for we divide a Line as we

please, by the Compass of Proportion: we do the Rule of Three without making use of Arithmetick: we extract the Square Root and Cube Root: we double the Cube: we measure all sorts of Triangles: we find the Content of Superficies, and the solidity of Bodies; we augment or diminish any figure whatever, according to what Proportion we please; and all those uses are Demonstrated by the foregoing Propositions.

PROPOSITION XIII. PROBLEM.

TO find a mean Proportional between Two Lines.

If you would have a mean Proportional between the Lines LV, VR: having joyned them together on a strait Line, divide the Line LR into two equal parts in the point M; and having descri∣bed a Semi-circle LTR on the Center M; draw the perpendicular VT, it shall be a mean Proportional between LV, VR. Draw the Lines LT, TR.

Demonstration. The Angle LTR, described in a Semi-circle, is right (by the 31st. of the 3d.) and (by the 8th.) the Triangles LVT, TVR, are like; there

is thence the same Reason in the Tri∣angle LVT, of LV to VT, as of VT to VR in the Triangle TVR, (by the 4th.) So then VT is a mean Proportional between LV and VR.

USE.

WE Reduce to a Square any Rect∣angular Parallelogram whatever by this Proposition. For example, in the Rectangle comprehended under LV, VR, I will demonstrate hereafter, that the Square of VT is equal to a Rectangle comprehended under LV, and VR.

PROPOSITION XIV. THEOREM.

EQuiangular and equal Parallelograms have their Sides reciprocal and equi∣angular Parallelograms whose sides are reci∣procal, are equal.

If the Parallelograms L and M be equi∣angular and equal, they shall have their sides reciprocal; that is to say, that there shall be the same Reason of CD to DE, as of FD to DB. For seeing they have their Angles equal, they may be joyned after such manner that their joyned

sides CD, DE, be on one streight Line (by the 15th. of the 1st.) continue the Sides AB, GE; you will have com∣pleated the Parallelogram BDEH.

Demonstration. Seeing the Paralle∣logram L and M are equal, they shall have the same reason to the Parallelogram BDEH. Now the reason of the pa∣rallelogram L, to the Parallelogram BDEH, is the same with that of the Base CD to the Base DE (by the 1st.) and that of the Parallelogram M, or DFGE, is the same with that of the Base FD to the Base BD. Thence there is the same reason of CD to DE, as of FD to BD.

Secondly, if the equiangular Paralle∣lograms L and M, have their Sides re∣ciprocal, they shall be equal.

Demonstration. The Sides of the Pa∣rallelograms are reciprocal; that is to say, that there is the same Reason of CD to DE, as of FD to BD: Now as the Base CD is to DE, so is the pa∣rallelogram L to the parallelogram BDEH, (by the first,) and as FD is to DB, so is the parallelogram M to BEDH; there is thence the same Rea∣son of L to BDEH, as of M to the same BDEH, so then (by the 7th. of

the 5th.) the parallelograms L, and M are equal.

PROPOSITION XV. THEOREM.

EQual Triangles which have one Angle equal, have the Sides which form that Angle reciprocal; and if their sides be re∣ciprocal, they shall be equal.

If the Triangles F and G, being equal, have their Angles ACB, ECD, equal: their sides about that Angle shall be reci∣procal; that is to say, that there shall be the same Ratio of BC to CE, as of CD to CA. Dispose the Triangles after such a manner, that the Sides CD, CA, be one straight Line: seeing the Angles ACB, ECD, are supposed equal, the Lines BC, CE, shall be also a straight Line, (by the 14th. of the 1st.) Draw the Line AE.

Demonstration. There is the same Ratio of the Triangle ABC to the Tri∣angle ACE, as of the Triangle ECD, equal to the first, to the same Triangle ACE, (by the 7th. of the 5th.) Now as ABC, is to ACE, so is the Base BC

to the Base CE, (by the 1st.) seeing they have the same vertical A; and as FCD is to ACE, so is the Base CD to CA. Now if it be supposed that the sides are reciprocal; that is to say, that there be the same Ratio of BC to CE, as of CD to CA; the Triangles ABC, CDE, shall be equal, because they would then have the same Ratio to the Triangle ACE.

PROPOSITION XVI. THEOREM.

IF four Lines be proportional, the Rect∣angle comprehended under the first and fourth, is equal to the Rectangle compre∣hended under the second and third, and if the Rectangle comprehended under the ex∣trems, be equal to the Rectangle comprehen∣ded under the means, then are the four Lines proportional.

if the Lines AB, CD, be proportional, that is to say, that there is the same Ratio of A to B, as of C to D; the Rectangle comprehended under the first A, and the fourth D, shall be equal to the Rectangle comprended under B and C.

Demonstration. The Rectangles have their Angles equal, seeing it is Right; they have also their Sides reciprocal; they are thence equal (by the 16th.)

In like manner, if they be equal, their sides are reciprocal; that is to say, that there is the same Ratio of A to B, as of C to D.

PROPOSITION XVII. THEOREM.

IF three Lines be proportional, the Rect∣angle comprehended under the first and third is equal to the Square of the mean, and if the Square of the mean be equal to the Rectangle of the extrems, the three Lines are then proportional.

If the three Lines A, B, D, be pro∣portional; the Rectangle comprehended under A, and under D, shall be equal to the Square of B. Take C equal to B; there shall be the same Ratio of A to B, as of C to D: thence the four Lines A, B, C, D, are proportional.

Demonstration. The Rectangle under A and D shall be equal to the Rectangle under B and C (by the foregoing:) Now this last Rectangle is a Square, seeing

the Lines B, and C, are equal: thence the Rectangle under A and D is equal to the Square of B.

In like manner, if the Rectangle under A and D be equal to the Square of B, there shall be the same Ratio of A to B, as of C to D; and seeing that B and C are equal, there shall be the same Ratio of A to B, as of B to D.

USE.

THose Four Propositions demonstrateth that Rule of Arithmetick which we commonly call the Rule of Three, and con∣sequently the Rule of Fellowship, false posi∣tion, and all other which are performed by Proportion. For example, let there be pro∣posed these three Numbers, A Eight, B six, C four, and it is required to find the fourth proportional. Suppose it to be found, and let it be D. The Rectangle comprehended under A and D is equal to to the Rectangle comprehended under B and C. Now I can have this Rectangle by Multiplying B by C; that is to say, six by four, and I shall have twenty four, thence the Rectangle comprehended under A and D, is twenty four, wherefore dividing the same by A

eight, the Quotient three is the number I look for.

PROPOSITION XVIII. THEOREM.

TO describe a Poligon like to another on a Line given.

There is proposed the Line AB, on which one would describe a Poligon like unto the Poligon CFDE. Having di∣vided the Poligon CFDE into Tri∣angles, make on the Line AB a Triangle ABH like unto the Triangle CFE; that is to say, make the Angle ABH equal to the Angle CFE, and BAH equal to FCE. So then the Triangles ABH, CFE, shall be equiangled (by the 32d. of the first.) make also on BH, a Triangle equiangled to FDE.

Demonstration. Seeing the Triangles which are parts of the Poligons, are equiangular, the two Poligons are equiangular. Moreover, seeing the Tri∣angles ABH, CFE, are equiangular, there is the same Ratio of AB to BH, as of CF to FE, (by the 4th.) In like manner, the Triangles HBG, EFD, being equiangular, there shall be the

same Ratio of BH to BG, as of FE to FD; and by equality there shall be the same Ratio of AB to BG, as of CF to FD. And so of the rest of the sides. Thence (by the first Definition,) the Poligons are like to each other.

USE.

IT is on this Proposition we establish the greatest part of the practical ways to take the plane of a place, of an Edifice, of a Field, of a Forrest, or of a whole Country; for making use of the equal parts of a Line for Feet or for Chains; we de∣scribe a figure like unto the Prototype, but lesser, in which we may see the Proportion of all its Lines. And because it is easier on paper than on the ground; we may compre∣hend in this Proposition all Geodes•…•…, all Chorography, all Geographical Charts, and ways of reducing of the greater into a lesser; wherefore this Proposition extends almost to all Arts, in which it is necessary to take a design or Model.

PROPOSITION XIX. THEOREM.

LIke Triangles are in duplicate Ratio to their homologous sides.

If the Triangles ABC, DEF, be like or equiangular; they shall be in duplicate Ratio of their homologous Sides BC, EF; that is to say, that the Ratio of the Triangle ABC, to the Triangle DEF shall be in duplicate Ratio of BC to EF; wherefore, by seeking the third Proportional HI to the Lines BC, EF; or so making it, that there may be the same Ratio of BC to EF, as of EF to HI; the Triangle ABC shall have the same Ratio to the Triangle DEF, as the Line BC hath to the Line HI. Which is called a du∣plicate Ratio, or doubled Reason (by the 11th. Def. of the fifth.) Let BG and HI be equal; and let the Line AG be drawn.

Demonstration. The Angle B and E of the Triangles ABG, DEF, are equal: also seeing the Triangle ABC, DEF, are like; there shall be the same Ratio (by the 4th.) of AB to DE, as

of BC to EF: Now as BC is to EF, so is EF to HI, or BG: thence as AB is to DE, so is EF to BG: and con∣sequently the sides of the Triangles ABG, DEF, being reciprocal, the Triangles shall be equal (by the 15th.) Now (by the first,) the Triangle ABC hath the same Ratio to the Triangle ABG, as BC to BG, or HI: thence the Triangle ABC hath the same Ratio to the Triangle DEF, as BC to HI.

USE.

THOSE Propositions do correct the opinions of many, who easily imagine, that like figures have the same Ratio as their Sides. For example, let there be proposed two Squares, Two Penta∣gons, two Hexagons, two Circles, and let the sides of the first be double to the side of the second; the first figure shall be quadruple to the second. If the side of the first be triple to that of the second, the first figure shall be nine times greater than the second. So that to find a Square triple to another, there must be sought a mean Proportional between one and three, which would be almost 1¾ for the side of the tripple figure.

PROPOSITION XX. THEOREM.

LIke Poligons may be divided into so many like Triangles, and they are in duplicate Ratio to their homologous sides.

If the Poligons ABCDE, GHIML, be like; they may be divided into so many like Triangles, and which shall be like parts of their whole. Draw the Lines AC, AD, GI, GL.

Demonstration. Seeing the Poligons are alike, their Angles B and H shall be equal, and there shall be the same Ratio of AB to BC, as of GH to HI, (by the 15th.) thence the Triangles ABC, GHI are alike: and (by the 4th.) there shall be the same Ratio of BC to CA, as of HI to GI. Moreover, seeng there is the same Ratio of CD to BC, as of IL to IH; and the same Ratio of BC to CA, as of HI to GI. There shall be by equality the same Ratio of CD to CA, as of IL to GI. Now the Angles BCD, and HIL, being equal, if you take away the equal Angles ACB, GIH, the Angles ACD, GIL, shall be equal. Thence the Triangles ACD

GIL, shall be like (by the 15th.) So that it is easie after the same manner to go round about the Angles of the Polygons, and to prove they are circular or alike.

I further add, that the Triangles are in the same Ratio as are the Polygons.

Demonstration. Seeing all the Tri∣angles are like, their sides shall be in the same Ratio (by the 4th.) Now each Tri∣angle is to its like, in duplicate Ratio of its homologous sides (by the 19th.) thence each Triangle of one Polygon, to each Triangle of the other Polygon, is in duplicate Ratio to the sides; which being the same; there shall be the same Ratio of each Triangle to its like, as of all the Triangles of one Polygon, to all the Triangles of the other Polygon (by the 12th. of the 5th.) that is to say, as of the one Poligon to the other.

Coroll. 1. Like Polygons are in dupli∣cate Ratio to their sides.

Coroll. 2. If three Lines are continually Proportional, the Polygon described on the first, shall have the same Ratio to the Polygon described on the second, as the first hath to the third; that is to say, in duplicate Ratio of that of the first Line to the second.

USE from the Learned Dr. Barrow.

BY this is found a method of enlarging or diminishing a Right Lined figure in a Ratio given, as if you would have a Pentagon Quintuple of that Pentagon whereof CD is the side; then betwixt AB and 5 AB, find out a mean Propor∣tional; upon this raise a Pentagon like unto that given, and it shall be quintuple also of the Pentagon given.

Hence also, if the Homologous sides of like figures be known, then will the Propor∣tion of the figure be evident, viz. by finding out a third Proportional.

PROPOSITION XXI. THEOREM.

POlygons which are like to a Third Poly∣gon, are also like one to the other.

If Two Polygons are like to a third, they shall be like one to the other; for each of them may be divided into as many like Triangles as there is in the third. Now the Triangles which are like unto the same Third, are also like one to the other, because the Angles which are

equal to a Third, are equal one to the other; and the Angles of the Triangles being equal, those of the Polygons of whom it is compounded, are so like∣wise.

I farther add that the sides of Tri∣angles being in the same Ratio, those of the Polygons shall be so likewise, seeing they are the same.

PROPOSITION XXII. THEOREM.

LIke Polygons described on four Lines which are Proportional, are also Pro∣portional. And if the Polygons be in the same Ratio, the Lines on which they are described shall be so also.

If there be the same Ratio of BC to EF, as of HT to MN; there shall also be the same Ratio of the Polygon ABC to the like Polygon DEF, as of the Polygon HL to the like Polygon MO. Seek to the Lines BC, EF, a third Proportional G; and to the Lines HT, MN, a third Proportional P (by the 11th.) seeing there is the same Ratio of BC to EF, as of HT to MN; and

of EF to G, as of MN to P: there shall be by equality the same Ratio of BC to G, as of HT to P; and this Ratio shall be duplicate or doubled of the Ratio of BC to EF, or of HT to MN.

Demonstration. The Polygon ABC to the Polygon DEF, is in duplicate or doubled Ratio of the Ratio of BC to EF (by the 21st.) that is to say, as BC is to G; and the Polygon HL to MO, hath the same Ratio as HT to P. There is therefore the same Ratio of ABC to DEF, as of HL to MO.

And if like Polygons are proportional; the Lines being in sub-duplicate Ratio, shall be also Proportional.

USE.

A,	B,	C,	D.
3.	2.	6.	4.
9.	4.	36.	16
E,	F,	G,	H.

THis Proposition may be easily applied to Numbers. If the Numbers A, B, C, D, are Propor∣tional, their Squares EF, GH, shall be so likewise; which we make use of in Arithmetick, and yet more use thereof in Algebra.

PROPOSITION XXIII. THEOREM.

EQuiangular Parallelograms have their Ratio compounded of the Ratio of their sides.

If the Parallelograms L and M be equi∣angular; the Ratio of L to M shall be compounded of that of AB to DE, and that of DB to DF. Joyn the Paralle∣lograms in such a manner that their sides BD, DF, be on a streight Line, as also CD, DE; which may be, if they be equiangular. Compleat the Paralle∣logram BDEH.

Demonstration. The Parallelogram L hath the same Ratio to the Parallelo∣gram BDEH, as the Base AB hath to the Base BH or DE (by the first.) the Parallelogram BDEH hath the same Ratio to the Parallelogram DFGE, that is to say M, as the Base BD hath to the Base DF. Now the Ratio of the Parallelogram L to the Parallelogram M, is compounded of that of L to the Pa∣rallelogram BDEH, and that of BDEH to the Parallelogram M. Thence the

Ratio of L to M, is compounded of that of AB to DE, and of that of BD to EG. For example, if AB be eight, and BH five, BD four, DF seven; say as four is to seven, so is five to eight ¾; you shall have these three numbers eight, five, eight ¾, eight to five shall be the Ratio of the Parallelogram L to BDEH, the same as that of AB to DE; five to eight ¾ shall be that of the Parallelogram BDEH to M. So then taking away the middle term which is five, you shall have the Ratio of eight to eight ••, for the Ratio compounded of both; or as 4 times 8 or 32, to 5 times 7 or 35.

PROPOSITION XXIV. THEOREM.

IN all sorts of Parallelograms; those through which the Diameter passeth, are like to the greater.

Let the Diameter of the Parallelo∣gram AC pass through the Parallelo∣grams EF, GH: I say they are like unto the Parallelogram AC.

Demonstration. The Parallelograms AC, EF, have the same Angle B; and because in the Triangles BCD, IF, is

Parallel to the Base DC, the Triangles BFI, BCD, are equiangular. There is therefore (by the 4th.) the same Ratio of BC to CD, as of BF to FI; and consequently the sides are in the same Ratio. In like manner, IH is Parallel to BC; there shall then be the same Ratio of DH to HI, as of DC to BC; and the Angles are also equal, all the sides being Parallel: thence (by the 8th. Def.) the Parallelograms EF, GH, are like unto the Parallelogram AC.

USE.

I Made use of this Proposition in the Tenth Proposition of the last Book of Perspective, to shew that an Image was drawn like unto the Original, with the help of a Parallelogram Composed of four Lines.

PROPOSITION XXV. PROBLEM.

TO describe a Polygon like unto a given Polygon, and equal to any other Right Lined figure.

If you would describe a Polygon equal

to the Right Lined figure A, and like unto the Polygon B; make a Parallelo∣gram CE equal to the Polygon B, (by the 34th. of the first.) and on DE, make a Parallelogram EF equal to the Right Lined figure A, (by the 45th. of the first.) Then seek a mean Proportional GH, between CD and DF (by the 13th.) Lastly, make on GH a Polygon O, like unto B (by the 18th.) It shall be equal to the Right Lined figure A.

Demonstration. Seeing that CD, GH, DF, are continually Proportional, the Right Lined figure B described on the first, shall be to the Right Lined figure O described on the second, as CD is to DF, (by the Coroll. of the 20th.) Now as CD is to DF, so is the Parallelogram CE to the Parallelogram EF, or as B to A, seeing they are equal. There is thence the same Ratio of B to O, as of B to A. So then (by the 7th. of the 5th.) A and O are equal.

USE

THis Proposition contains a change of figures, keeping always the same Area, which is very useful, principally in Practical Geometry, to reduce an irregular figure into a Square.

PROPOSITION XXVI. THEOREM.

IF in one of the Angles of a Parallelo∣gram, there be described a lesser Pa∣rallelogram like unto the greater, the Dia∣meter of the greater shall meet the Angle of the lesser.

If in the Angle D, of the Parallelo∣gram AC; be described another lesser DG, like thereto: The Diameter DB shall pass through the point G. For if it did not, but it passed through I, as doth the Line BID. Draw the Line IE Parallel to HD.

Demonstration. The Parallelogram DI, is like unto the Parallelogram AC, (by the 24th.) Now it is supposed that the Parallelogram DG is also like there∣to; thence the Parallelograms DI, DG, would be like; which is impossible: otherwise, there would be the same Ratio of HI to IE or GF, as of HG to GF; and (by the 7th. of the 5th.) the Lines HI, HG, would be equal.

The Twenty Seventh, Twenty Eighth, and Twenty Ninth Propositions are unnecessary.

PROPOSITION XXX. PROBLEM.

TO cut a Line given into extream and mean Proportion.

It is proposed to cut the Line AB in extream, and mean Proportion; that is to say, in such a manner, that there may be the same Ratio of AB to AC, as of AC to CB. Divide the Line AB (by the 11th. of the second) in such a manner that the Rectangle comprehended under AB, CB, be equal to the Square of AC.

Demonstration. Seeing the Rectangle of AB, BC, is equal to the Square of AC; there will be the same Ratio of AB to AC, as of AC to BC (by the 17th.)

USE.

THis Proposition is necessary in the Thirteenth Book of Euclid, to find the length of the Sides of some of the five Regular Bodies. Father Lucas of St. Se∣pulchres

hath Composed a Book of the Pro∣prieties of a Line, which is cut into ex∣tream and mean Proportion.

PROPOSITION XXXI. THEOREM.

A Polygon which is described on the Base of a Right Angled Triangle, is equal to the other Two like Polygon descri∣bed on the other Sides of the same Triangle.

If the Triangle ABC hath a Right Angle BAC; the Polygon D described on the Base BC, is equal to the like Po∣lygons F, and E, described on the Sides AB, AC.

Demonstration. The Polygons D, E, F, are amongst themselves in duplicate Ratio of their homologous Sides BC, AC, AB, (by the 20th.) If there were described a Square on those sides, they would be also in duplicate Ratio to their sides. Now (by the 47th. of the first.) the Square of BC would be equal to the Squares of AC, AB; thence the Poly∣ligon D described on BC, is equal to the like Polygons E and F, described on AB, AC.

USE.

THis Proposition is made use of to make all sorts of figures greater or lesser, for it is more universal then (the 47th. of the 1st.) which notwithstanding is so useful, that it seemeth that almost all Geometry is established on this Principle.

The 32d. Proposition is unnecessary.

PROPOSITION XXXIII. THEOREM.

IN equal Circles, the Angles as well those at the Center as those at the Cir∣cumference, as also the Sectors are in the same Ratio as the Arks, which serve to them as Base.

If the Circles ANC, DOF, be equal; there shall be the same Ratio of the Angle ABC to the Angle DEF, as of the Ark AC to the Ark DF. Let the Ark AG, GH, HC, he equal Arks, and consequently Aliquot parts of the Ark AC, and let be divided

the Ark DF, into so many equal to AG, as may be found therein, and let the Lines EI, EK, and the rest be drawn.

Demonstration. All the Angles ABG, GBH, HBC, DEI, IEK, and the rest, are equal (by the 37th. of the third;) so then AG an Aliquot part of AC, is found in the Ark DF, as many times as the Angle ABG an Aliquot part of the Angle ABC, is found in DEF; there is therefore the same Ratio of the Ark AC to the Ark DF, as of the Angle ABC to the Angle DEF. And because N and O are the halfs of the Angles ABC, DEF, they shall be in the same Ratio as are those Angles; there is therefore the same Ratio of the Angle N to the Angle O, as of the Ark AC to the Ark DF.

It is the same with Sectors; for if the Lines AG, GH, HC, DI, IK, and the rest were drawn, they would be equal (by the 28th. of the Third;) and each Sector would be divided into a Triangle and a Segment. The Triangles would be equal (by the 8th. of the first;) and the little Segments would be also equal (by the 24th. of the third;) thence all those little

Sectors would be equal: and so as many as the Ark BF containeth of Aliquot parts of the Ark of AC, so many the Sector DKF would contain Aliquot parts of the Sector AGC. There is therefore the same Reason of Ark to Ark, as of Sector to Sector.

Notes

• * 1.1

  Plate VI.
• * 1.2

  Fig. I.