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.

* 1.1 A Strait Line cannot have one of its parts in a Plane, and the other with∣out it.

If the Line AB be in the Plane AD, it being continued, shall not go without, but all its parts shall be in the same Plane. For if it could be that BC were a part of AB continued. Draw in the Plane CD, the Line BD, perpendi∣cular to AB: draw also in the same Plane, BE perpendicular to BD.

Demonstration. The Angles ABD, BDE, are both Right Angles; thence (by the 14th. of the first,) AB, BE, do make but one Line; and consequently BC, is not a part of the Line AB con∣tinued; otherwise two strait Lines CB, EB, would have the same part AB: that is AB would be part of both: which we have rejected as false in the Thirteenth Maxim of the first Book.

USE.

WE establish on this Proposition a principle in Gnomonicks, to

prove that the shadow of the stile falleth not without the Plane of a great Circle, in which the Sun is. Seeing that the end or top of the stile is taken for the Center of the Heavens; and consequently for the Center of all the great Circles: the shadow being always in a streight Line, with the Ray drawn from the Sun to the Opaque Body; this Ray being in any great Circle, the shadow must also be therein.

PROPOSITION II. THEOREM.

LInes which cut one another, are in the same Plane, as well as all the parts of a Triangle.

If the Two Lines BE, CD, cut one another in the Point A; and if there be made a Triangle by drawing the Base BC: I say that all the parts of the Tri∣angle ABC, are in the same plane, and that the Lines BE, CD, are likewise therein.

Demonstration. It cannot be said that any one part of the Triangle ABC, is in a Plane, and that the other part is without; without saying that one part of

a Line is in one Plane, and that the other part of the same Line is not therein; which is contrary to the first Proposition: and seeing that the sides of the Triangle are in th same Plane wherein the Tri∣angle is; the Lines BE, CD, shall be in the same Plane.

USE.

THis Proposition doth sufficiently de∣termine a Plane, by two streight Lines mutually intersecting each other, or by a Triangle; I have made use thereof in Opticks, to prove that the objective parallel Lines which fall on the Tablet, ought to be Represented by Lines which concur in a Point.

PROPOSITION III. THEOREM.

THe common section of two Places is a streight Line.

If Two Planes AB, CD, cut one another, their common section EF shall be a streight Line. For if it were not, take Two Points common to both Planes which let be E and F; and draw a

strait Line from the point E to the point F, in the Plane AB, which let be EHF. Draw also in the Plane CD a streight Line from E to F; if it be not the same with the former, let it be EGF.

Demonstration. Those Lines drawn in the Two Planes are two different Lines, and they comprehend a space; whch is contrary to the Twelfth Maxim. Thence they are but one Line, which being in both Planes, shall be their common section.

USE.

THis Proposition is fundamental. We do suppose it in Gnomonicks, when we represent in a Dial, the Circles of the hours, marking only the common section of their Planes, and that of the Wall.

PROPOSITION IV. THEOREM.

IF a Line be perpendicular to two other Lines which cut one another, it shall be also perpendicular to the Plane of those Lines.

If the Line AB be perpendicular to

the Lines CD, EF, which cut one ano∣ther in the point B; in such manner that the Angles ABC, ABD, ABE, ABF, be right, which a flat figure cannot re∣present; it shall be perpendicular to the Plane CD, EF; that is to say, that it shall be Perpendicular to all the Lines that are drawn in that Plane through the point B: as to the Line GBH. Let equal Lines be cut BC, BD, BE, BF; and let be drawn the Lines EC, DF, AC, AD, AE, AF, AG, and AH.

Demonstration. The four Triangles ABC, ABD, ABE, ABF, have their Angles Right in the Point B; and the Sides BC, BD, BE, BF, equal with the side AB common to them all. Therefore their Bases AC, AD, AE, AF, are equal (by the 4th. of the 1st.)

2. The Triangles EBC, DBF, shall be equal in every respect, having the Sides BC, BD, BE, BF, equal; and the Angles CBE, DBF, opposite at the vertex being equal; so then the Angles BCE, BDF, BEC, BFD, shall be equal (by the 4th. of the first,) and their Bases EC, DF, equal.

3. The Triangles GBC, DBH, having their opposite Angles CBG, DBH, equal, as also the Angles BDH,

BCG, and the sides BC, BD, they shall then have (by the 26th. of the 1st.) their Sides BG, BH, CG, DH, equal.

4. The Triangles ACE, AFD, having their sides AC, AD, AE, AF, equal, and the Bases EC, DF, equal, they shall have (by the 8th. of the 1st.) the Angles ADF, ACE, equal.

5. The Triangles ACG, ADH, have the Sides AC, AD, CG, DH, equal; with the Angles ADH, AGC: Thence they shall have their Bases AG, AH, equal.

Lastly, the Triangles ABH, ABG, have all their sides equal; thence (by the 27th. of the 1st.) the Angles ABG, ABH, shall be equal, and the Line AB perpendicular to GH. So then the Line AB shall be perpendicular to any Line which may be drawn through the point B, in the Plane of the Lines CD, EF, which I call perpendicular to the Plane.

USE.

THis Proposition cometh often in use in the first Book of Theodosius; for ex∣ample, to Demonstrate that the Axis of the World

is perpendicular to the Plane of the Equi∣noctial. In like manner in Gnomonicks, we Demonstrate by this Proposition, that the Equinoctial is perpendicular to the Me∣ridian in Horizontal Dials, it is not less useful in other Treatises; as in that of Astrolabes, or in the sections of Stone.

PROPOSITION V. THEOREM.

IF a Line be Perpendicular to three other Lines which cut one the other in the same Point, these three Lines shall be in the same Plane.

If the Line AB be perpendicular to three Lines BC, BD, BE, which cut one another in the point B, the Lines BC, BD, BE, are in the same plane. Let the plane AE be the Plane of the Lines AB, BE; and let CF be the Plane of the Lines BC, BD. If BE was the common section of the Two Planes, BE would be in the Plane of the Lines BC, BD, as we pretend it should. Now if BE be not the common section; let it be BG.

Demonstration. AB is perpendicular to the Lines BC, BD; it is then perpendi∣cular to the Plane CF, (by the 4th. and 5th. Def.) AB shall be also perpendicular to BG. Now it is supposed that it is perpendicular to BE; thence the Angles ABE, ABG, would be right, and equal, and notwithstanding the one is a part of the other. So then the Two Planes cannot have any other common section besides BE; it is therefore in the Plane CF.

PROPOSITION VI. THEOREM.

THe Lines which are perpendicular to the same Plane, are parallel.

If the Lines AB, CD, be perpendi∣cular to the same plane EF; they shall be parallel. It is evident that the inter∣nal Angles ABD, BDC, are right; but that is not sufficient, for we must also prove that the Lines AB, CD, are in the same Plane. Draw DG, perpendi∣cular to BD, and equal to AB: Draw also the Lines BG, AG, AD.

Demonstration. The Triangles ABD, BDG, have the sides AB, DG, equal: BD is common; the Angle•• ABD, BDG, are right. Thence the Bases AD, BG, are equal (by the 4th. of the 1st.) Moreover the Triangles ABG, ADG, have all their sides equal; thence the Angles ABG, ADG, are equal; and ABG being right, seeing AB is perpendicular to the Plane, the Angle ADG is right. Therefore the Line DG is perpendicular to the three Lines CD, DA, DB, which by consequence are in the same Plane (by the 5th.) Now the Line AB is also in the plane of the Lines AD, DC, (by the second;) thence AB, CD, are in the same plane.

Coroll. Two Lines which are parallel are in the same Plane.

USE.

WE demonstrate by this Proposition, that the hour Lines are parallels amongst themselves, in all Planes which are parallel to the Axis of the World; as in the Polar and Meridian Dials, and others.

PROPOSITION VII. THEOREM.

A Line which is drawn from one pa∣rallel to another, is in the same Plane which those Lines are.

The Line CB being drawn from the point B of the Line AB, to the point C of its parallel CD: I say that the Line CB is in the Plane of the Lines AB, CD.

Demonstration. The parallels AB, CD, are in the same Plane; in which if you draw a strait Line from the point C to the point B, it will be the same with CB; otherwise two strait Lines would inclose a space, contrary to the 12th. Maxim.

PROPOSITION VIII. THEOREM.

IF of Two parallel Lines, the one be perpendicular to a Plane, the other shall be so likewise.

As if of the Two parallel Lines AB, CD; the one AB be perpendicular to

the Plane EF: CD shall be so likewise. Draw the Line BD, seeing that ABD is a right Angle, and that the Lines AB, CD are supposed parallel; the Angle CDB shall be right (by the 30th. of the 1st.) Wherefore if I Demonstrate that the Angle CDG is Right, I shall prove (by the 4th.) that CD is perpendicular to the Plane EF. Make the right Angle BDG, and make DG equal to AB, then draw the Lines PG, AG.

Demonstration. The Triangles ABD, BDG, have their sides AB, DG, equal; the side BD is common, the Angles ABD, BDG, are right: Thence (by the 4th. of the 1st) the Bases AD, BG, are equal. The Triangles ADG, ABG, have all their sides equal; so then (by the 6th.) the Angles ADG, ABG, are equal. This last is right, seeing the Line AB is suppo∣sed perpendicular to the Plane EF; thence the Angle ADG, is right; and the Line DG being perpendicular to the Lines DB, DA, it shall be perpen∣dicular to the Plane of the Lines AD, DB; which is the same in which are the parallels AB, CD. So then the Angle GDC is a Right Angle (by the 5th. Def.)

PROPOSITION IX. THEOREM.

THe Lines which are Parallel to a third, are parallel amongst them∣selves.

If the Lines AB, CD, are parallel to the Line EF; they shall be parallel to each other, although they be not all three in the same Plane. Draw in the Plane of the Lines AB, EF, the Line HG perpendicular to AB, in shall be also perpendicular to EF, (by the 29th. of the first.) In like manner draw in the Plane of the Lines EF, CD, the Line HI perpendicular to EF, CD.

Demonstration. The Line EH being perpendicular to the Lines HG, HI; is also perpendicular to the Planes of the Lines HG, HI, (by the 4th.) thence (by the 8th.) the Lines AG, CI, are so likewise, and (by the 6th.) they shall be parallel.

USE.

THis Proposition serveth very often in Perspective, to determine on the

Plane the Image of the parallel Lines, and in the sections of stones, where we prove that the sides of the Squares are pa∣rallel to each other, because they are so to some Line in a different Plane. In Gno∣monicks we prove that vertical Circles ought to be drawn by perpendiculars on the Wall, because that the Lines which are their common section with the Wall, are parallel to the Line drawn from the Zenith to the Nadir.

PROPOSITION X. THEOREM.

IF Two Lines which meet in a point are parallel to Two other Lines in a different Plane; they will form an equal Angle.

If the Lines AB, CD; AE, CF, are parallel, although they be not all four in the same Plane; the Angles BAE, DCF, shall be equal. Let the Lines AB, CD; AE, CF, be equal: and draw the Lines BE, DF, AC, BD, EF.

Demonstration. The Lines AB, CD, are supposed parallel and equal; thence (by the 33d. of the first,) the Lines AC,

BD, are parallel and equal; as also AC, and EF; and (by the preceding) BD, EF, shall be also parallel and equal; and (by the 33d. of the first,) BE, DF, shall be parallel and equal. So then the Triangles BAE, DCF, have all their sides equal; and (by the 8th.) the Angles BAE, DCF, shall be equal.

Coroll. One might make some Propo∣sitions like unto this, which would not be unuseful; as this. If one should draw in a parallel Plane, the Line CD parallel to AB, and if the Angles BAC, DCF, are equal; the Lines AC, CF, are parallel.

USE.

WE Demonstrate by this Proposi∣tion, that the Angles which are made by the hour Circles in a Plane pa∣rallel to the Aequator, are equal to those which they make in the Plane of the Ae∣quator.

PROPOSITION XI. PROBLEM.

TO draw a perpendicular to a Plane, from a point given without the Plane.

If you would from the point C, draw a perpendicular to the Plane AB; draw the Line EF at discretion, and CF per∣pendicular thereto (by the 12th. of the first.) then draw (by the 11th. of the first,) in the Plane AB, the Line FG perpen∣dicular to ED, and CG perpendicular to FG. I demonstrate that CG shall be perpendicular to the Plane AB. Draw GH parallel to EF.

Demonstration. The Line EF being perpendicular to the Lines CF, FG, it shall be perpendicular to the Plane CFG (by the 4th.) and HG being pa∣rallel to EF, shall be also perpendicular to the same Plane (by the 8th.) and seeing that CG is perpendicular to the Lines GF, GH, it shall be perpendicular to the Plane AB, (by the 4th.)

PROPOSITION XII. PROBLEM.

TO draw a perpendicular to a Plane, from a point of the same Plane.

To draw from the point C a perpen∣dicular to the Plane AB; draw from the point E taken at discretion without the Plane, ED perpendicular to the same Plane (by the 11th.) Draw also (by the 30th. of the first,) CF parallel to DE, CF shall be perpendicular to the Plane AB, (by the 8th.)

PROPOSITION XIII. THEOREM.

THere cannot be drawn from the same point Two Perpendiculars to a Plane.

If the Two Lines CD, CE, drawn from the same point C, were perpendi∣cular to the Plain AB; and that CE was the common section of the Plane of those Lines, with the Plane AB; the Angles

ECF, DCF, would be Right; which is impossible.

I further add, that there cannot be drawn from the same point D Two per∣pendiculars DC, DF, to the same Plane AB: for having drawn the Line CF, there will be made Two Right Angles DCF, DFC, in one Triangle contrary to the 32d. of the first.

USE.

THis Proposition is necessary to shew, that a Line which is drawn perpendi∣cular to a Plane is well determined, seeing there can be drawn but only one from a Point.

PROPOSITION XIV. THEOREM.

PLanes are parallel to which the same Line is perpendicular.

If the Line AB be perpendicular to the Planes AC, BD, they shall be pa∣rallel; that is to say, they shall be in all places equidistant. Draw the Line DC

parallel to AB, (by the 30th. of the first,) and joyn the Lines BD, AC.

Demonstration. It is supposed that AB is perpendicular to the Planes AC, BD; thence the Line CD, which is parallel thereto, shall be also perpendi∣cular to them (by the 8th;) so then the Angles B and D, A and C; shall be Right; and (by the 29th. of the first,) the Lines AC, BD, shall be parallel. The figure ABCD shall be thence a pa∣rallelogram. Now (by the 34th. of the first,) the Lines AB, CD, are equal; that is to say, that the Planes in the Points A and B, C and D, are equi-distant: So then because we can draw the Line CD from any point whatever, the Planes AC, BD, shall be equi-distant in all places.

USE.

THeodosius Demonstrateth by this, that the Circles which have the same Poles, as the Equinoctial, and the Tropicks, are parallel; because the Axis of the World is perpendicular to their Planes.

PROPOSITION XV. THEOREM.

IF Two Lines which meet in the same point are parallel to Two Lines of ano∣ther plane; the Planes of those Lines shall be parallel.

If the Lines AB, AC, in the one Plane, are parallel to the Lines DF, DE, which are in another Plane; the Planes BC, FE, are parallel. Draw AI perpendicular to the Plane BC (by the 11th.) and GI, IH, parallel to FD, DE; they shall also be parallel to the Lines AB, AC, (by the 9th.)

Demonstration. The Lines AB, GI, are parallel, and the Angle IAB is Right, seeing AI is perpendicular to the Plane BC: thence (by the 29th. of the first) the Angle AIG is Right. AIH is also Right: Thence (by the 4th.) the Line AI is perpendicular to the Plane GH; and it being also perpendicular to the Plane BC, the Planes BC, FE, shall be parallel (by the 14th.)

PROPOSITION XVI. THEOREM.

IF a Plane cut Two Planes which are pa∣rallel, its common sections with them shall be parallel Lines.

If the Plane AB cutteth two other parallel Planes AC, BD: I demonstrate that the common sections AF, BE, shall be parallels. For if they were not, they would meet if continued, for ex∣ample in the point G.

Demonstration. The Lines AF, BE, are in the Planes AC, BD, and do not go out of the same (by the first;) where∣fore if they meet each other in G, the Planes would meet each other like wise, and consequently they would not be pa∣rallel, contrary to what we have sup∣posed.

USE.

WE Demonstrate by this Proposition, in the Treatise of Conick sections and Cylindricks, that the Cone or Cylinder being cut by a parallel to its Base, the sections are Circles; we describe Astrolabes:

We prove in Gnomonicks, that the Angles the hour Circles make with a Plane parallel to a great Circle, are equal to those which they make with the Circle it self; we demon∣strate in Perspective, that the Image of the objective Lines which are perpendicular to the Tablet, concur in the point of vision.

PROPOSITION XVII. THEOREM.

TWo Lines are divided proportionally by parallel Planes.

Let the Lines AB, CD, be divided by parallel Planes; I say that AE, EB, and CF, FD, are in the same Ratio. Draw the Line AD, which meeteth the Plane EF in the point G: draw also AC, BD, FG, EG.

Demonstration. The Plane of the Tri∣angle ABD, cutteth the three Planes; thence (by the 16th) the sections BD, EG, shall be parallel; and (by the 2d. of the sixth,) there shall be the same. Ratio of AE to EB, as of AG to GD. In like manner, the Triangle ADC cutteth the Planes EF, AC; thence the sections AC, GF, are parallel; and there shall be the same Ratio of FC to FD, as of

AG to GD, that is to say, as of AE to EB.

PROPOSITION XVIII. THEOREM.

IF a Line be perpendicular to a Plane, all the Planes in which that Line shall be found, shall be perpendicular to the same Plane.

If the Line AB be perpendicular to the Plane ED; all the Plains in which it shall be found, shall be also perpen∣dicular to ED. Let AB be in the Plane AE, which hath for common section with the Plane ED, the Line BE, unto which let there be drawn the perpendicu∣lar FI.

Demonstration. The Angles ABI, BIF, are Right; thence the Lines AB, FI, shall be parallel; and (by the 8th.) FI shall be perpendicular to the Plane ED. So then the Plane AE shall be perpendicular to the Plane ED, (by the 5th. Def.)

USE

THe first Proposition in Gnomonicks, and which may stand for a fundamen∣tal, is established on this Proposition, it is also often made use of in Spherical Trigo∣nometry, in perspective and generally in all Treatises, which are obliged to consider se∣veral Planes.

PROPOSITION XIX. THEOREM.

IF Two Planes which cut each other, are perpendicular to another, their common section shall also be perpendicular thereto.

If the Two P anes AB, ED, cut each other, and are perpendicular to the Plane IK; there common section EF is perpendicular to the Plane IK.

Demonstration. If EF be not per∣pendicular to the Plane IK; let there be drawn in the Plane AB, the Line GF perpendicular to the common section BF; and seeing the Plane AB is perpen∣dicular to the Plane IK; the Line GF, shall be perpendicular to the same Plane.

Let there also be drawn FH perpendi∣cular to the common section DF, it shall be perpendicular to the Plane IK. Thus we shall have Two perpendiculars to the same Plane, drawn from the same point F, contrary to the 13th. Proposition. We must therefore conclude that EF is perpendicular to the Plane IK.

USE.

WE demonstrate by this Proposition, that the Circle which passeth through the Poles of the World, and through the Zenith, is the Meridian, and cutteth in two equally all the Diurnal Arks, and that every star employes the same space of time from their rising to this Circle, as they do from this Circle to their setting.

PROPOSITION XX. THEOREM.

IF three Plain Angles compose a solid Angle, any Two of them shall be greater than the third.

If the Angles BAC, BAD, CAD, Compose the solid Angle A; and if BAC be the greatest of them; the other Two

BAD, CAD, taken together, are greater than BAC. Let CAE be equal to CAD, and let the Lines AD, AE, be equal. Draw the Lines CEB, CD, BD.

Demonstration. The Triangles CAE, CAD, have the sides AD, AE, equal; CA, common; and the Angles CAD, CAE, equal; thence (by the 4th. of the first,) the Bases CD, CE, are equal. Now the Sides CD, BD, are greater than the side BC (by the 20th of the first,) wherefore taking away the equal Lines CD, CE; the Line BD shall be greater than BE. Moreover the Triangles BAE, BAD, have their sides AD, AE, equal; AB common; and the Base BD greater than EB: thence (by the 18th. of the first;) the Angle BAD is greater than the Angle BAE; and adding the Angles CAD, CAE; the Angles BAD, CAD, shall together be greater than BAE, CAE, that is to say, than CAB.

PROPOSITION XXI. PROBLEM.

ALL the Plain Angles which Compose a solid Angle are less than four Right.

If the Plane Angles BAC, CAD, BAD, Compose the solid Angle A; they shall be less than four Right. Draw the Lines BC, CD, BD, and you shall have a Pyramid, which hath for Base the Triangle BCD.

Demonstration. There is made a solid Angle at the point B, of which the Angles ABC, ABD, taken together, are greater than the Angle CBD at the Base. In like manner ACB, ACD, are greater than BCD; and the Angles ADC, ADB, are greater than CDB. Now all the Angles at the Base CDB, are equal to Two Right; thence the Angles ABC, ABD, ACB, ACD, ADC, ADB, are greater than Two Right. And because all the Angles of the Three Triangles BAC, BAD, CAD, are equal to six Right; in taking away more than Two Right,

there will remain less than four Right for the Angles which are made at the point A. If the solid Angle A was Composed of more than Three plain Angles; in such manner that the Base of the Pyramid was a Polygon; it might be divided into Triangles: and by making the supputa∣tion, one would always find that all the Plane Angles which form the solid Angle, taken together, are less than four Right.

USE.

THose Two Propositions serve to deter∣mine, when of several plain Angles there may be made a solid Angle, which is often necessary in the Treatise of cutting of stone, and in the following Propositions.

The Two and Twentieth and Three and Twentieth Propositions are needless.

PROPOSITION XXIV. THEOREM.

IF a solid Body be termined by parallel Planes, the opposite Planes shall be like Parallelograms, and equal.

If the solid AB be terminated by pa∣rallel

Planes, the opposite Superficies shall be like and equal parallelograms.

Demonstration. The parallel Planes AC, BE, are cut by the Plane FE; wherefore the common sections AF, DE, are parallel (by the 15th.) In like manner DF, AE, whence AD shall be a parallelogram. I demonstrate after the same manner, that AG, FB, CG, and the rest, are parallelograms. I fur∣ther add, that the opposite parallelo∣grams; for example AG, BF, are like and equal. The Lines AE, EG, are parallel to the Lines FD, BD, and also equal; thence the Angles AEG, FDB, are equal (by the 15th.) I may also demonstrate that all the sides, and all the Angles of the opposite parallelo∣grams are equal. Thence the Parallelo∣grams are like and equal.

PROPOSITION XXV. THEOREM.

IF a parallelepipedon be divided by a Plane parallel to one of its sides; the Two solids which result from this Division, shall be in the same Ratio as their Bases.

If the parallelepipedon AB be divi∣ded

by the Plane CD parallel to the op∣posite Planes AF, BE, the solid AC to BD, shall be in the same Ratio as the Base AI to the Base DG. Let us ima∣gine that the Line AH, which repre∣sents the height of the figure is divided into as many parts as one pleaseth; for example in Ten Thousand, which we may take to be indivisible, that is to say, without thinking that they can be divi∣ded. Let us also suppose or imagine as many Superficies parallel to the Base AG, as there is parts in the Line AH: I ex∣press but one for all, as OS; in which manner the Solid AB may be Composed of all those Surfaces of equal thickness, as would be a Ream of Paper composed of all its Sheets, when laid on each other. It is evident that then the Solid AC shall be composed of Ten Thousand Surfaces equal to the Base AI, (by the preceeding;) and the Solid BD shall contain Ten Thousand Superficies equal to the Base DG.

Demonstration. Each Surface of the Solid AC, hath the same Ratio to each Surface of the Solid BD, as the Base AI hath to the Base DG; seeing each of them are equal to their Bases; thence (by the 12th. of the 5th.) all the Super∣ficies

of the Solid AC taken together, shall have the same Ratio to all the Sur∣faces of the Solid DB; as hath the Base AI to the Base DG. Now all the Sur∣faces of the Solid AC, compose the Solid AC, which hath no other parts but its Surfaces, and all the Surfaces of the Solid DB, are nothing else but the Solid DB; thence the Solid AC to the Solid DB, hath the same Ratio as hath the Base AI to the Base DG.

USE

THis way of Demonstration is that of Cavallerius: I find it to be very clear, provided it be made use of as it ought, and that the Line which is the mea∣sure of the thickness of its surfaces, be taken in the same manner in the one and in the other Term. I will make use thereof hereafter, to make more easie some Demon∣strations which otherwise are too incombred.

PROPOSITION XXVI. THEOREM.

A Parallelepipedon is divided into Two equally, by a Diagonal Plane.

Let the Parallelepipedon AB be divi∣ded

by the Plane CD, drawn from one Angle to the other: I say that it divideth the same into Two equally Let the Line EA, be supposed to be divided into as many parts as one listeth; and that from each, one hath drawn as many Planes parallel to the Base AF; each of those Planes is a Parallelogram equal to the Base AF (by the 24th.)

Demonstration. All the Parallelograms which can be drawn parallel to the Base AF, are divided into Two equally by the Plane CD; for the Triangle which will be made on each side of the Plane CD, have their common Base equal to CG; and their sides equal, seeing they are those of a parallelogram. Now it is evident that the Parallelepipedon AB, containeth nothing else but its parallelo∣gram Surfaces, each of which is divided into Two equal Triangles: therefore the parallelepipedon AB, is divided into Two equally by the Plane CD.

The Twenty Seventh and Twenty Eighth Propopositions are useless, according to this way of Demonstration.

PROPOSIT. XXIX. XXX. & XXXI THEOREM.

PArallelepipedons having the same height and the same, or equal Bases, are equal.

If the Parallelepipedon CD, AB, have equal perpendicular height AE, FG, with equal Bases AH, CI, or the same, they shall be equal. Let be put the Two Bases AH, CI, on the same Plane, seeing their heights are equal, the Bases EB, FD, shall be in the same Plane, parallel to that of the Bases AH, CI. Let it be imagined that the perpendicular FG or EA is divided into as many equal parts as one listeth; for example in Ten Thousand; and there be drawn through each Surfaces or Planes of the same thickness; to express which I mention but one for all, which shall be KM. Each Surface shall form in each Solid a Plane parallel like and equal its Base (by the 24th) as KL, OM, and there shall be as many of them in the one as in the other; seeing their thickness which I take perpendicularly in the Line of their height, is equal.

Demonstration. There is the same Ratio of the Base AH to the Base CI, as of each plain KL to OM: Now there is the same number of them in the one as there is in the other, so then (by the 15th. of the fifth) there shall be the same Ratio of all the Antecedents to all the Consequents, that is to say of all the Solid AB to all the Solid CD; as of the Base AH to the Base CI. Now it is supposed that the Bases are equal: thence the Solids AB, CD, are equal.

Coroll. To find the Solidity of a Pa∣rallelepipedon, we Multiply the Base by its height taken perpendicularly, because that perpendicular sheweth how many times their is found Surfaces equal to the Base. As if I take a Foot for an indivi∣sible measure, that is to say, which I will not conceive subdivided: If the Base should contain Twelve Foot Square, and that the perpendicular height were Ten Foot; I should have One Hundred and Twenty Cubick Feet, for the Soli∣dity of the Body AB. For seeing the height AE, is Ten Foot; I may make Ten, parallelograms equal to the Base, and which shall have each of them one Foot thickness. Now the Base with one Foot thickness, makes Twelve

Foot Cubick; it shall then contain One Hundred and Twenty, if it hath Ten Foot height.

PROPOSITION XXXII. THEOREM.

PArallelepipedons of the same height are in the same Ratio as their Bases.

I have Demonstrated this Proposition in the precedent, in proving that there is the same Ratio of the Parallepipedon AB to the Parallelepipedon CD, as of the Base AH to the Base CI.

Coroll. Parallelepipedons which have equal Bases, are in the same Ratio as are their height; as the Parallelepipe∣dons AB, AL, which have for their Perpendicular height AK, AE; for if the height AK were divided into as many Aliquot parts as one would, and AE into as many equal parts of the first, as it would contain, and if there were drawn through each as many Planes pa∣rallel to the Base, I say as many as AE would contain of Aliquot parts of AK, so many Solids AB shall contain of Sur∣faces equal to the Base, which are Aliquot

parts of the Solid AL. Therefore (by the 5th. of the 5th.) there shall be the same Ratio of the Solid AB to the Solid AL, as of the height to the height AK.

USE

THe Three precedent Propositions con∣tain almost all the mensuration of Pa∣rallelepipedons, and serve as the first prin∣ciples in this matter. It is thus we measure the Solidity of a Wall, by Multiplying their Bases by their heights.

PROPOSITION XXXIII. THEOREM.

LIke Parallelepipedons are in tripled Pro∣portion of their Homologus sides.

If the Parallelepipedons AB, CD, are like; that is to say, if all the Planes of the one are like unto all the Planes of the other; and if all their Angles are equal, in such manner that they may be placed in a strait Line, that is to say, that AE, EF; HE, EI; GE, EC; be strait Lines, and that there is the same Ratio of AE to EF, as of HE to EI,

and of GE to EC. I demonstrate that four Solids are continually proportio∣nals, according to the Ratio of EA, to that which is homologous thereto, which is EF or DI.

Demonstration. The Parallelepipe∣don AB, hatth the same Ratio to EL of the same height, as hath the Base AH to the Base EO (by the 32d.) Now the Base AH to the Base EO, hath the same Ratio as AE to EF, (by the 1st. of the 6th.) In like manner the Solid EL to the Solid EK, is the same as that of the Base EO to the Base ED, that is to say, as of HE to EI. In fine the Solid EK to the Solid EN, hath the same Ratio, as hath the height GE to the height EC, (by the coroll:) or (taking EF for their common height,) as of the Base GI to the Base CI, that is to say, as of GE to EC. Now the Ratio of AE to EF, and of HE to EI, and of GE to EC, is the same, as we did suppose: By consequence, there is the same Ratio of AB to EL, as of EL to EK, and of EK to CD. Therefore (by the 11th. Def. of the 5th.) the Ratio of AB to CD, shall be triple of that of AB to EL, or of AE to his homologous EF.

Coroll. If four lines are continually proportional, the Parallelepipedon de∣scribed on the first, hath the same Ratio to a like Parallelepipedon described on the second, as the first hath to the fourth, for the Ratio of the first to the fourth is tripled of that of the first to the second.

USE.

YOu may comprehend by this Proposition, that the Celebrated Proposition of the Duplication of the Cube, proposed by the Oracle, consisteth in this to find two means continually Proportional. For if you pro∣pose for this first term, the side of the first Cube; and that the fourth term be the double of this first: If you find Two mean proportionals, the Cube described on the first Line shall have the same Ratio to that described on the second, as the first Line hath to the fourth, which shall be as One to Two. We also correct by this Proposition the false opinion of those which imagine, that like Solids are in the same Ratio as are their sides, as if a Cube of one foot long were the half of a Cube of Two Foot long, although it be but the eighth part thereof. This is the Principle of the Calibres, which

may be applyed not only to Cannon Shot, but also to all like Bodies. For example, I have seen a person which would make a Naval Architecture, and would keep the same Proportions in all sorts of Vessels: but argued thus; if a Ship of One Hundred Tuns, ought to be Fifty Foot long, a Ship of Two Hundred ought to be One Hundred Foot by the Keel. In which he was much mistaken, for instead of making a Ship of twice that Burden, he would make one Octuple. He ought to have given to the Second Ship to be double in Burden to the First, somewhat less than Sixty Three Foot.

PROPOSITION XXXIV. THEOREM.

EQual Parallelepipedons have their Bases and height Reciprocal, and those which have their Bases and height Recipro∣cal, are equal.

If the Parallepipedons AB, CD, are equal, they shall have their Bases and height reciprocal; that is to say, there shall be the same Ratio of the Base AE to the Base CF, as of the height CH to the height AG. Having made

CI equal to AG, draw the Plane IK parallel to the Base CF.

Demonstration. The Parallelepipedon AB, hath the same Ratio to CK of the same height, as hath the Base AE to CF, (by the 32d.) Now as AB to CK, so is CD to the same CK, seeing that AB and CD are equal; and as CD is to CK, which hath the same Base, so is the height CH to the height CI (by the Coroll. of the 32d.) Wherefore, as the Base AE is to the Base CF, so is the height CH to the height CI or AG.

I further add, that if there be the same Ratio of AE to CF, as of the height CH to the height AG; the Solids AB, CD, shall be equal.

Demonstration. There is the same Ratio of AB to CK, of the same height, as of the Base AE to the Base CF, (by the 32d) there is also the same Ratio of the height CH to the height CI or AG, as of CD to CK; we suppose that the Ratio of AE to CF, is the same of that of CH to CI or AG; so then there shall be the same Ratio of the Solid AB to the Solid CK, as of the Solid CD to the same Solid CK. Therefore (by the 8th. of the 5th.) the Solids AB, CD, are equal.

USE.

THis reciprocation of Bases, ren∣dreth these Solids easie to be mea∣sured; it hath also a kind of Analogy with the Sixteenth Proposition of the Sixth Book which beareth this, that parallelo∣grams equiangled and equal, have their Sides reciprocal, and it Demonstrateth also as well the practice of the Rule of Three.

PROPOSITION XXXVI. THEOREM.

IF Three Lines are continually propor∣tional, the Solid Parallelepipedon made of them is equal to an equiangular paralle∣lepipedon, which hath all it sides equal to the middle Line.

If the Lines A, B, C, be continually proportional, the parallelepipedon FE made of them, that is to say, which hath the side FI equal to the Line A, and HE equal to B, HI equal to C; is equal to the equiangular Parallelepipe∣don KL, which hath its sides LM, MN, KN,

equal to the Line B. Let there be drawn from the Points H and N, the Lines HP, NQ, perpendicular to the Planes of the Bases; they shall be equal, seeing that the Solid Angles E and K are supposed equal, (in such manner that if they did penetrate each other, they would not surpass each other) and that the Lines EH, KN, are supposed equal. There∣fore the heights HP, NQ, are equal.

Demonstration. There is the same Ratio of A to B, or of FI to LM, as of B to C, or of LM to HI; so then the parallelogram FH comprehended under FI, IH, is equal to the paralle∣logram LN, comprehended under LM, MN, equal to B (by the 15th. of the 6th.) the Bases are thence equal. Now the heights HP, NQ, are so likewise; therefore (by the 31st.) the Parallelepi∣pedons are equal.

PROPOSITION XXXVII. THEOREM.

IF four Lines are proportional, the Paral∣lepipedons described on those Lines shall be Proportional, and if like Parallelepipe∣dons are Proportional, their homologous sides shall be so likewise.

If A to B bein the same Ratio as C to D; the like Parallepipedons which shall have for their homologous sides A, B, C, D, shall be in the same Ratio.

Demonstration. The Parallelepipedon A to the Parallelepipedon B, is in triple Ratio of that of the Line A to the Line B, or of that of the Line C to the Line D. Now the Parallelepipedon C to the Parallelepipedon D, is also in triple Ratio of that of C to D, (by the 33d.) there is therefore the same Ratio of the Parallepipedon A to the Parallepipedon B, as of the Parallelepipedon C to the Parallelepipedon D.

PROPOSITION XXXVIII. THEOREM.

IF Two Planes be Perpendicular to each other; the Perpendicular drawn from one Point of one of the Planes to the other, shall fall on the common section.

If the Planes AB, CD, being perpen∣dicular to each other, there be drawn from the Point E of the Plane AB, a Line perpendicular to the Plane CD; it shall fall on AG, the common section

of the Planes. Draw EF perpendicular to the common section AG.

Demonstration. The Line EF per∣pendicular to AG, the common section of the Planes, which we supposed per∣pendicular, shall be perpendicular to the Plane CD (by the 3d. Def.) and seeing there cannot be drawn from the Point E two perpendiculars to the Plane CD (by the 13th.) the perpendicular shall fall on the common Section AG.

USE.

THis Proposition ought to have been after the 17th. seeing it hath respect to Solids in general. It is of use in the Treatise of Astrolabes, to prove that in the Analemma, all the Circles which are perpendicular to the Meridian, are strait Lines.

PROPOSITION XXXIX. THEOREM.

IF there be drawn in a parallelepipedon Two Planes dividing the opposite sides equally; their common Section and the Dia∣meter doth also cut each other equally.

Let the opposite sides of the parallele∣pipedon AB, be divided into Two equally by the Planes CD, EF; their common section GH, and the Diameter BA shall divide each other equally in O. Draw the Lines BG, GK, AH, LH. I prove in the first place, that they make but one strait Line; for the Triangles DBG, KMG, have their sides BD, KM, equal, seeing they are the halves of equal sides; as also GD, GM. Moreover, DB, KM, being parallel, the Alternate Angles BDG, GMK, shall be equal (by the 29th. of the first;) so then (by the 4th. of the first,) the Triangles DBG, KGM, shall be equal in every respect; and conse∣quently the Angles BGD, KGM. Now (by the 16th. of the 1st.) BG, GK, is but one Line, as also LH, HA; thence AL, BK, is but one Plane, in which is found the Diameter AB and GH, the common section of the Planes. The Plane AL, BK, cutting the parallel Planes AN, CD, shall have its com∣mon sections GH, AK, parallel; and (by the 4th. of the 6th.) there shall be the same Ratio of BK to GK, as of BA to BO; and of AK to OG, (by the 4th. of the 6th.) Now BK is double to

BG; thence BA is double to BO; as AK equal to HG, is double to GO. So then the Lines GH, AB, cut each other equally in the Point O.

Coroll. 1. All the Diameters are di∣vided in the Point O.

Coroll. 2. Here may be put Corollaries which depend on several Propositions; for example, that Triangular Prisms of equal height, are in the same Ratio as are their Bases; for the Parallelepipedons of which they are the halves (by the 32d.) are in the same Ratio as their Bases; so that the half Bases and the half pa∣rallelepipedons, that is to say the Prisms, shall be in the same Ratio.

Coroll. 3. Polygon Prisms of the same height, are in the same Ratio as are their Bases; seeing they may be reduced to Triangular Prisms, which shall each of them be in the same Ratio as their Bases.

Coroll. 4. There may be applied to the Prism the other Propositions of the Parallelepipedons; for example, that equal Prisms have their heights and Bases Reciprocal, that like Prisms are in triple Ratio of their homologous sides.

USE.

THis Proposition may serve to find the Center of gravity of Parallelepipe∣dons, and to Demonstrate some Propositions of the Thirteenth and Fourteenth Books of Euclid.

PROPOSITION XL. THEOREM.

THat Prism which hath for its Base a Parallelogram double to the Trian∣gular Base of another, and of the same height, is equal thereto.

Let there be proposed Two Triangu∣lar Prisms ABE, CDG, of the same height; and let be taken for the Base of the first the Parallelogram AE, double to the Triangle FGC, the Base of the second Prism. I say that those Prisms are equal. Imagine that the Parallelepi∣pedons AH, GI, are drawn.

Demonstration. It is supposed that the Base AE is double to the Triangle FGC: Now the Parallelogram GK being also double to the same Triangle

(by the 34th. of the 1st.) the Parallelo∣grams AE, GK, are equal: and con∣sequently the Parallelepipedons AH, GI, which have their Bases and heights equal, are equal; and Prisms which are the halves (by the 28th.) shall be also equal.

Notes

• * 1.1

  Plate VII. Prop. I.