the two Sides FD, DE; but the Angle BAC is equal to the Angle FDE. Therefore the Base BC is* equal to the Bafe EF, the Triangle to the Tri- * 4. 1. angle, and the other Angles to the other Angles. Wherefore the Angle ACB is equal to the Angle DFE; but the Right Angle ACK is equal to the Right Angle DFN; and therefore the remaining Angle BCK is equal to the remaining Angle EF N. For the fame Reason, the Angle CBK is equal to the Angle FEN; and fo becaufe BCK, EFN, are two Triangles, having two Angles equal to two Angles, each to each, and one Side equal to one Side, which is at the equal Angles, viz, BC equal to EF; therefore they fhall have the other Sides equal to the other Sides. Therefore CK is equal to FN, but AC is equal to DF. Therefore the two Sides AC, CK, are equal to the two Sides DF, FN, and they con-tain Right Angles; confequently the Bafe AK is equal to the Bafe DN. And fince A H is equal to DM, the Square of A H fhall be equal to the Square of DM; but the Squares of A K, KH, are equal to the Square of A H; for the Angle AKH is a Right Angle, and the Squares DN, NM, are equal to the Square of DM, fince the Angle DNM is a Right Angle. Therefore the Squares of AK, KH, are equal to the Squares of DN, NM; of which the Square of A K is equal to the Square of DN. Wherefore the Square of KH remaining, is equal to the remaining Square of NM; and fo the Right Line HK is equal to MN. And fince the two Sides HA, AK, are equal to the two Sides MD, DN, each to each, and the Bafe HK has been proved equal to the Bafe NM, the Angle HAK shall be † equal to the Angle † 8. 1 MDN; which was to be demonftrated.

Coroll. From hence it is manifeft, that if there be two Right-lined plane Angles equal, from whofe Points equal Right Lines be elevated on the Planes of the Angles, containing equal Angles with the Lines first given, each to each; Perpendiculars drawn from the extreme Points of thofe elevated Lines to the Planes of the Angles first given, are equal to one another.

† 14. 6.

PROPOSITION XXXVI.

THEOREM.

If three Right Lines be proportional, the folid Parallelepipedon made of them, is equal to the folid Parallelepipedon made of the Middle Line, if it be an equilateral one, and equiangular to the aforefaid Parallelepipedon.

I

ET three Right Lines A, B, C, be proportional, viz. Let A be to B, as B is to C. I fay, the Solid made of A, B, C, is equal to the equilateral Solid made of B, equiangular to that made on A, B, C.

a

Let E be a folid Angle contained under the three plane Angles DEG, GEF, FED; and make DE, GE, EF, each equal to B, and complete the folid Parallelepipedon EK. Again, put LM equal to A, 26 of this, and at the Point L, at the Right Line LM, make* folid Angle contained under the plane Angles NLX, XLM, MLN, equal to the folid Angle E; and make LN equal to B, and LX to C. Then becaufe A is to B, as B is to C, and A is equal to L M, and B to LN, EF, EG, or ED, and C to LX; it shall be as LM is to EF, fo is GE to LX. And fo the Sides about the equal Angles MLX, GEF, are reciprocally proportional. Wherefore the Parallelogram MX is equal to the Parallelogram GF. And fince the two plane Angles GEF, XLM, are equal, and the Right Lines LN, ED, being equal, are erected at the angular Points containing equal Angles with the Lines firft given, each to each; the Perpendiculars 1 Cor. 35. of drawn from the Points ND, to the Planes drawn. thro' XLM, GEF, are equal one to another. Therefore the Solids LH, EK, have the fame Altitude; but folid Parallelepipedons that have equal Bafes, and the * 31 of this. fame Altitude, are equal to each other. Therefore the Solid HL is equal to the Solid EK. But the Solid HL is that made of the three Right Lines A, B, C, and the Solid EK that made of the Right Line B. Therefore, if three Right Lines be proportional, the folid Parallelepipedon made of them, is equal to the folid Parallelepipedon made of the middle Line, if it be an

this.

equilateral one, and equiangular to the aforefaid Parallelepipedon; which was to be demonftrated.

PROPOSITION XXXVII.

THEOREM.

If four Right Lines be proportional, the folid Parallelepipedons fimilar, and in like manner described from them, fhall be proportional. And if the folid Parallelepipedons, being fimilar, and alike defcribed, be proportional, then the Right Lines they are defcribed from, fhall be proportional.

ET the four Right Lines AB, CD, EF, GH, be proportional, viz, let AB be to CD, as EF is to GH, and let the fimilar and alike fituate Parallelepipedons KA, LC, ME, NG, be defcribed from them. I fay, KA is to LC, as ME is to NG.

*

33 of this

For because the folid Parallelepipedon KA is fimilar to LC, therefore KA to LC fhall have a Pro- * portion triplicate of that which AB has to CD. For the fame Reason, the Solid ME to NG will have a triplicate Proportion of that which EF has to GH, But AB is to CD, as EF is to GH. Therefore AK is to LC, as ME is to N G. And if the Solid AK be to the Solid LC, as the Solid ME is to the Solid N G. I fay, as the Right Line AB is to the Right Line CD, fo is the Right Line EF to the Right Line GH. For because AK to LC has + a Proportion triplicate of † 33 of this. that which AB has to CD, and ME to NG has a Proportion triplicate of that which EF has to GH, and fince AK is to LC, as ME is to NG; it fhall be as A B is to CD, fo is EF to GH. Therefore, if four Right Lines be proportional, the folid Parallelepipedons fimilar, and in like Manner defcribed from them, fhall be proportional. And if the folid Parallelepipedons, being fimilar and alike defcribed, be proportional, then the Right Lines they are defcribed from fhall be proportional; which was to be demonftrated.

Def. 4 of this.

this.

If

PROPOSITION XXXVIII.

THEOREM.

a Plane be perpendicular to a Plane, and a Line be drawn from a Point in one of the Planes perpendicular to the other Plane, that Perpendicular fall fall in the common Section of the Planes.

ET the Plane CD be perpendicular to the Plane

fome Point E be taken in the Plane CD. I fay, a Perpendicular, drawn from the Point E to the Plane AB, falls on AD.

*

For if it does not, let it fall without the fame, as EF meeting the Plane A B in the Point F, and from the Point Flet FG be drawn in the Plane AB perpendicular to AD; this fhall be perpendicular to the Plane CD; and join E G. Then because FG is perpendicular to the Plane CD, and the Right Line EG in the Plane of CD touches it: The Angle + Def. 3. of FGE fhall be + a Right Angle. But EF is alfo at Right Angles to the Plane AB; therefore the Angle EFG is a Right Angle. And fo two Angles of the Triangle EFG are equal to two Right Angles; which is abfurd. Wherefore if a Right Line, drawn from the Point E, perpendicular to the Plane AB, does not fall without the Right Line AD: And so it must neceffarily fall on it. Therefore, if a Plane be perpendicular to a Plane, and a Line be drawn from a Point in one of the Planes perpendicular to the other Plane, that Perpendicular shall fall in the common Section of the Planes; which was to be demonftrated.

PRO

PROPOSITION XXXIX.

THEORE M.

If the Sides of the oppofite Planes of a folid Parallelepipedon be divided into two equal Parts, and Planes be drawn thro' their Sections; the common Section of the Planes, and the Diameter of the folid Parallelepipedon, Shall divide each other into two equal Parts.

LE

ET the Sides of CF, AH, the oppofite Planes of the folid Parallelepipedon AF, be cut in half in the Points K, L, M, N, X, O, P, R, and let the Planes KN, X R, be drawn through the Sections: Alfo let YS be the common Section of the Planes, and DG the Diameter of the folid Parallelepipedon. · I fay, YS, DG, bifect each other, that is, YT is equal to TS, and DT to TG.

For join DY, YE, BS, SG. Then because DX is parallel to OE, the alternate Angles DXY, YOE are* equal to one another. And because DX is equal * 29. 1 to OE, and YX to YO, and they contain equal Angles, the Bafe DY fhall be + equal to the Bafe YE; † 4 1. and the Triangle DXY to the Triangle YOE, and the other Angles equal to the other Angles: Therefore the Angle XYD is equal to the Angle OYE; and fo DYE is a Right Line. For the fame Rea¬† 14. 1. fon BSG is alfo a Right Line, and BS is equal to SG. Then because CA is equal and parallel to DB, as alfo to EG, DB fhall be equal and parallel to EG; and the Right Lines DE, GB join them: Therefore DE is parallel to BG, and D, Y, G, S are Points * taken in each of them, and DG, YS are joined. Therefore DG, YS are † in one Plane. And fince DE is † 7 of this. parallel to BG, the Angle EDT fhall be equal to * 29. 1. the Angle BGT, for they are alternate. But the Angle DTY, is equal to the Angle GTS. Therefore‡ 15. 1. DTY, GTS are two Triangles, having two Angles of the one equal to two Angles of the other, as likewife one Side of the one equal to one Side of the other, viz. the Side DY equal to the Side GS: For they are Halves of DE, BG: Therefore they shall

*

have