125 of this. is the Bafe M P to the Bafe PT, and fo is MC to CT: Therefore, as the Bafe EH is to the Bafe NP, fo is MC to CT. But CT is equal to A G; wherefore, as the Base E H is to the Base NP, fo is MC to AG: Therefore, the Bafes and Altitudes of the equal folid Parallelepipedons A B, CD, are reciprocally proportional.

Now, let the Bafes and Altitudes of the folid Parallelepipedons A B, CD, be reciprocally proportional; that is, let the Base E H be to the Bafe NP, as the Altitude of the Solid CD is to the Altitude of the Solid A B. I fay, the Solid A B is equal to the Solid CD.

For, let again the infiftent Lines be at Right Angles to the Bafes; then, if the Bafe EH be equal to the Bafe NP, and EH is to NP as the Altitude of the Solid CD is to the Altitude of the Solid A B; the Altitude of the Solid CD fhall be equal to the Altitude of the Solid A B. But folid Parallelepipedons, that stand upon equal Bases, and have the fame Alti33 of this. tude, are equal to each other; therefore the Solid A B is equal to the Solid CD.

*

But now let the Base EH not be equal to the Bafe N P, and let E H be the greater; then the Altitude of the Solid CD is greater than the Altitude of the Solid AB; that is, CM is greater than AG: Again, put CT equal to AG, and compleat the Solid C V, as before; and then, because the Bafe EH is to the Bafe NP, as MC is to A G, and AG is equal to CT; it fhall be, as the Bafe E H is to the Bafe N P, fo is MC to CT: But as the Base EH is to the Base N P, so is the Solid A B to the Solid CV; for the SolidsA B,CV, have equal Altitudes; and as MC is to CT, fo is the Bafe MP to the Bafe PT, and fo the Solid CD to the Solid CV: Therefore as the Solid A B is to the Solid CV, fo is the Solid CD to the Solid CV: But fince each of the Solids A B, CD, has the fame Proportion to CV; the Solid A B fhall be equal to the Solid CD; whence, the two folid Parallelepipedons A B, CD, whofe Bafes and Altitudes are reciprocally proportional, are equal; which was to be demonftrated.

Now, let the infiftent Lines F E, BL, GA, KH, XN, DO, M C, RP, not be at Right Angles to the Bafes; and from the Points F, G, B, K, X, M, D, R, let there be drawn Perpendiculars to the Planes

of

of the Bafes EH, NP, meeting the fame in the Points S, T, Y, V, Q, Z, n, o, and compleat the Solids FV, X. Then, I fay, if the Solids A B, CD, be equal, their Bafes and Altitudes are reciprocally proportional; viz. as the Bafe EH is to the Bafe NP, fo is the Altitude of the Solid CD to the Altitude of the Solid A B.

For, because the Solid A B is equal to the Solid CD, and the Solid A B is equal to the Solid B T; for they * 30 of this, ftand upon the faine Bafe F K, and have the fame Altitude; and the Solid D C is equal to the Solid D Z, fince they stand upon the fame Bale X R, and have the fame Altitude; therefore the Solid B T fhall be equal to the Solid DZ. But the Bafes and Altitudes of thofe equal Solids, whofe Altitudes are at Right Angles to their Bales, are + reciprocally proportional; therefore as+ From the Bafe F K is to the Base X R, fo is the Altitude of what bas the Solid D Z to the Altitude of the Solid BT. But been before the Bafe F K is equal to the Bafe EH, and the Bafe XR to the Bale NP; wherefore, as the Bafe EH is to the Bafe NP, fo is the Altitude of the Solid DZ to the Altitude of the Solid B T. But the Solids D Z, DC, have the fame Altitude, and fo have the Solids BT, BA; therefore the Bafe EH is to the Bale NP, as the Altitude of the Solid DC is to the Altitude of the Solid A B; and fo, the Bafes and Altitudes of equal folid Parallelepipedons are reciprocally proportional.

proved.

Again, let the Bafes and Altitudes of the folid Parallelepipedons A B, CD, be reciprocally proportional; viz. as the Bafe EH is to the Bafe N P, fo let the Altitude of the Solid CD be to the Altitude of the Solid A B: I fay, the Solid A B is equal to the Solid CD.

For, the fame Construction remaining, because the Bafe E H is to the Bafe NP, as the Altitude of the Solid CD is to the Altitude of the Solid A B; and fince the Base E H is equal to the Bafe FK, and NP to XR; it fhall be, as the Base F K is to the Bafe X R, fo is the Altitude of the CD to the Altitude of the Solid A B. But the Aes of the Solids A B, B T, are the fame; as alfo of the Solids CD, DZ; therefore the Base F K is to the Bafe X R, as the Altitude of the Solid D Z is to the Altitude of the Solid BT; wherefore the Bafes and Altitudes of the folid Parallelepipe

Q3

dons

+ From

bat has

been becre

proved.

1

dons BT, DZ, are reciprocally proportional; but thofe folid Parallelepipedons, whofe Altitudes are at. Right Angles to their Bafes, and the Bases and Altitudes are reciprocally proportional, are equal to + each other. But the Solid BT is equal to the Solid BA, for they stand upon the fame Bafe FK, and have the fame Altitude; and the Solid DZ is alfo equal to the Solid DC, fince they ftand upon the fame Base XR, and have the fame Altitude: Therefore the Solid A B is equal to the Solid CD; whence folid Parallelepipedons, whofe Bafes and Altitudes are reciprocally proportional, are equal; which was to be demonftrated.

PROPOSITION XXXV.

THEOREM.

If there be two plane Angles equal, and from the Vertices of thofe Angles two Right Lines be elevated above the Planes, in which the Angles are, containing equal Angles with the Lines firft given, each to its correfpondent one; and if in thofe elevated Lines any Points be taken, from which Lines be drawn perpendicular to the Planes in which the Angles first given are, and Right Lines be drawn to the Angles first given from the Points made by the Perpendiculars in the Planes; thofe Right Lines will contain equal Angles with the elevated Lines.

LE

ET BAC, EDF,be two equal Right-lined plane Angles, and from A and D, the Vertices of those Angles, let two Right Lines, A G and DM, be elevated above the Planes of the faid Angles, making equal Angles with the Lines first given, each to its correfpondent one; viz. the Angle M DE equal to the Angle GA B, and the Angle MDF to the Angle GAC; and take any Points G and M in the Right Lines AG, DM; from which let GL and MN be drawn perpendicular to the Planes paffing thro' BA C, EDF, meeting the fame in the Points L and N; and join LA and ND. I fay, the Angle G A L is equal to the Angle MDN.

Make

Make A H equal to D M, and thro' H let HK be drawn parallel to GL; but G L is perpendicalar to the Plane paffing thro' BAC; therefore HK fhall be ++8 of this. alfo perpendicular to the Plane paffing thro' BAC: Draw from the Points K and N, to the Right Lines A B, AC, DE, and D F, the Perpendiculars K B,K C, NE, NF; and join H C, CB, MF, FE: Then, because the Square of HA is equal to the Squares of 47. I. HK, KA; and the Squares of KC and CA are + equal to the Square of K A; the Square of HA fhall be equal to the Squares of H K, KC, and CA: But the Square of HC is equal to the Squares of HK and KC; therefore the Square of H A will be equal to the Squares of H C and CA; and fo the Angle HCA is ‡‡ 48. 1. a Right Angle. For the fame Reafon, the Angle D FM is also a Right Angle; therefore the Angle ACH is equal to D F M: But the Angle HA C is alfo equal to the Angle MDF; therefore the two Triangles MDF, HAC, have two Angles of the one equal to two Angles of the other, each to each, and one Side of the one equal to one Side of the other; viz. that which is fubtended by one of the equal Angles; that is, the Side H A equal to DM; and fo the other Sides of the. one fhall be equal to the other Sides of the other,* 26. 1. each to each: Wherefore AC is equal to D F. In like manner we demonftrate, that A B is equal to DE: For, let H B, ME, be joined; then, because the Square of AH is equal to the Squares of AK and KH; and the Squares of AB and BK are equal to the Square of A K; the Squares of A B, BK, and K H, will be equal to the Square of AH. But the Square of B H is equal to the Squares of B K, KH; for the Angle HK B is a Right Angle, because HK is perpendicular to the Plane paffing thro' BAC; therefore the Square of A H is equal to the Squares of AB and BH: Wherefore the Angle ABH is † at 48. 1. Right Angle. For the fame Reafon the Angle DEM is alfo a Right Angle; and the Angle B AH is equal to the Angle EDM, for fo it is put; and A His equal to DM; therefore A B is also equal to DE:* And fo, fince AC is equal to DF, and A B to DE the two Sides CA, AB, fhall be equal to the two. Sides FD, DE: But the Angle B A C is equal to the Angle FDE; therefore the Bafe BC is equal to

*

*

*

4. I.

*

the Bafe EF, the Triangle to the Triangle, and the other Angles to the other Angles: Wherefore the Angle A C B is equal to the Angle DF E. But the Right Angle ACK is equal to the Right Angle D F Ñ; and therefore the remaining Angle BCK is equal to the remaining Angle EFN. For the fame Reason, the Angle CBK is equal to the Angle FEN; and fo, becaufe BCK and 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 C K is equal to F N. But A C is equal to DF; therefore the two Sides A C, CK, are equal to the two Sides D F, FN; and they contain Right Angles; confequently, the Bafe A K is equal to the Bafe D N. And fince AH is equal to DM, the Square of A H fhall be equal to the Square of D M: But the Squares of A K and KH are equal to the Square of A H; for the Angle A K H is a Right Angle; and the Squares of DN and N M are equal to the Square of DM, fince the Angle DN M is a Right Angle; therefore the Squares of AK and KH are equal to the Squares of DN and 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 N M ; and fo the Right Line H K is equal to M N. And fince the two Sides H A, A K, are equal to the two Sides MD, DN, each to each, and the Bafe KH has been proved equal to the Bafe NM, the Angle HA K, or GAL, fhall be + equal to the Angle MDN; which was to be demonflrated.

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 firft giyen, are equal to one another.

PRO