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PROPOSITION XXXIV.

THEOREM.

The Bafes and Altitudes of equal folid Parallelepipedons, are reciprocally proportional; and thofe folid Parallelepipedons, whofe Bafes and Altitudes are reciprocally proportional, are equal.

LE

ET AB, CD, be equal folid Parallelepipedons, I fay, their Bafes and Altitudes are reciprocally proportional, that is, as the Bafe EH is to the Bafe NP, fo is the Altitude of the Solid CD to the Alti tude of the Solid AB.

First, let the infiftent Lines AG, EF, LB, HK, CM, NX, OD, PR, be at Right Angles to their Bafes. I fay, as the Bafe EH is to the Bafe NP, fo is CM to AG. For if the Bafe EH be equal to the Base NP, and the Solid AB is equal to the Solid CD, the Altitude CM fhall alfo be equal to the Altitude AG: For if when the Bafes EH, NP, are equal, the Altitudes AG, CM, are not fo; then the Solid AB will not be equal to the Solid CD, but it is put equal to it. Therefore the Altitude CM is not unequal to the Altitude A G, and fo they are neceffarily equal to one another; and confequently, as the Base EH is to the Base NP, fo fhall CM be to AG. But now let the Base EH be unequal to the Bafe NP, and let EH be the greater: Then fince the Solid AB is equal to the Solid CD, CM is greater than AG; for otherwise it would follow that the Solids AB, CD, are not equal, which are put fuch. Therefore make CT equal to AG, and complete the folid Parallelepipedon VC upon the Bafe NP, having the Altitude CT. Then because the folid AB is equal to the folid CD, and VC is fome other Solid; and fince equal Magnitudes have the fame Propor tion to the fame Magnitude, it fhall be as the Solid AB is to the Solid CV, fo is the Solid CD to the Solid CV; but as the Solid A B is to the Solid CV, +32 of this, fo is the Base EH to the Bafe NP; for AB, CV, are Solids having equal Altitudes. And as the Solid $25 of this. CD is to the Solid CV, fo is ‡ the Bafe MP to the

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Bafe

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 AG. Wherefore as the Bafe EH is
to the Bafe NP, fo is MC to AG. Therefore the
Bafes and Altitudes of the equal folid Parallelepipedons
AB, CD, are reciprocally proportional.

Now, let the Bafes and Altitudes of the folid Parallelepipedons AB, CD, be reciprocally proportion al; that is, let the Bafe EH be to the Bafe NP, as the Altitude of the Solid CD is to the Altitude of the Solid AB. I fay, the Solid AB is equal to the Solid CD.

For let again the infiftent Lines be at Right Angles to the Bases; then if the Base EH be equal to the Base NP, and EH is to NP, as the Altitude of the Solid CD is to the Altitude of the Solid AB; the Altitude of the Solid CD fhall be equal to the Altitude of the Solid AB. But folid Parallelepipedons that ftand upon equal Bafes, and have the fame Altitude, are equal to each other. Therefore the Solid* 31 of this AB is equal to the Solid CD..

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But now let the Base EH not be equal to the Base NP, and let EH 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 complete the Solid CV as before: And then because the Base EH is to the Bafe NP, as MC is to AG, and AG is equal to CT; it fhall be as the Bafe EH is to the Bafe NP, fo is MC to CT; but as the Base EH is to the Base NP, fo is the Solid AB to the Solid VC; for the Solids AB, CV, have equal Altitudes : And as MC is to CT, fo is the Bafe MP to the Base PT, and fo the Solid CD to the Solid CV. Therefore as the Solid AB 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 AB fhall be equal to the Solid CD; which was to be demonftrated.

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

*

+ From what bas

Points S, T, Y, V, Q, Z, N, 9, and complete the Solids FV, X 2, Then, I fay, if the Solids AB, CD, be equal, their Bases and Altitudes are reciprocally proportional, viz. as the Bafe EH is to the Base NP, fo is the Altitude of the Solid CD to the Altitude of the Solid AB.

*

For because the Solid AB is equal to the Solid 30 of this. CD, and the Solid A B is equal to the Solid BT; for they ftand upon the fame Bafe, have the fame Altitude, and their infiftent Lines are not in the fame Right Lines, and the Solid DC is equal to the Solid DZ, fince they stand upon the fame Base, XR have the fame Altitude, and their infiftent Lines are not in the fame Right Lines; the Solid BT fhall be equal to the Solid DZ; but the Bafes and Altitudes of thofe equal Solids, whofe Altitudes are at Right Angles to their Bases, are + reciprocally proportional. Therefore as the Bafe FK is to the Bafe XR, fo is the Altitude of the Solid DZ, to the Altitude of the Solid BT; but the Bafe FK is equal to the Base EH, and the Base XR to the Bafe NP. Wherefore as the Base EH is to the Base NP, fo is the Altitude of the Solid DZ to the Altitude of the Solid BT; but the Solids DZ, DC, have the fame Altitude, and fo have the Solids BT, BA. Therefore the Base EH is to the Bafe NP, as the Altitude of the Solid DC is to the Altitude of the Solid AB; and fo the Bases and Altitudes of equal Solids are reciprocally proportional.

been before proved.

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

For the fame Conftruction remaining, because the Bafe EH is to the Bafe NP, as the Altitude of the Solid CD is to the Altitude of the Solid AB; and fince the Base EH is equal to the Bafe FK, and NP to XR, it fhall be as the Bafe FK is to the Base X R, fo is the Altitude of the Solid CD to the Altitude of the folid AB; but the Altitudes of the Solids AB, BT, are the fame; as alfo of the Solids CD, DZ. Therefore the Base FK is to the Bafe XR, as the

Altitude

Altitude of the Solid DZ is to the Altitude of the Solid BT; wherefore the Bafes and Altitudes of the folid Parallelepipedons BT, DZ, are reciprocally proportional; but thofe folid Parallelepipedons, whofe Altitudes are at Right Angles to their Bases, and the Bafes 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 Base FK, have the fame Altitude, and their infiftent Lines are not in the fame Right Lines; and the Solid DZ is alfo equal to the Solid DC, fince they stand upon the fame Bafe X R, have the fame Altitude, and their infiftent Lines are not in the fame Right Lines. Therefore the Solid A B is equal to the Solid CD; 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 first 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 firft given from the Points made by the Perpendiculars in the Planes, thofe Right Lines will contain equal Angles with the elevated Lines.

ET BAC, EDF, be two equal Right-lined L'Angles,

Angles; and from A, D, the Vertices of those Angles, let two Right Lines AG, DM, be elevated above the Planes of the faid Angles, making equal Angles with the Lines firft given, each to its correfpondent one, viz. the Angle MDE equal to the Angle GAB, 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, MN, be drawn perpendicular to the Planes paffing thro' BAC, EDF, meeting the fame in the Points L, N, and

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join

† 47.1.

48. 1.

join LA, ND. I fay the Angle GAL is equal to the Angle MDN.

Make A H equal to DM, and thro' H let, HK be drawn parallel to GL; but GL is perpendicular to the Plane paffing thro' BAC. Therefore H K shall *8 of this. be* alfo perpendicular to the Plane paffing thro' BAC. Draw from the Points K, N, to the Right Lines AB, AC, DE, DF, the Perpendiculars KB, KC, NE, NF, and join HC, CB, MF, FE. Then because the Square of HA is equal to the Squares of HK, KA, and the Squares of KC, CA, are † equal to the Square of KA; the Square of HA fhall be equal to the Squares of HK, KC, CA; but the Square of HC is equal to the Squares of HK, KC. Therefore the Square of HA will be equal to the Squares of HC and CA; and fo the Angle HCA is a Right Angle. For the fame Reason, the Angle DFM is alfo a Right Angle. Therefore the Angle ACH is equal to DFM; but the Angle HAC 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 HA equal to DM; and fo the other Sides of the one, fhall be equal to the other Sides of the other, each to each. Wherefore AC is equal to DF. In like manner we demonftrate, that AB is equal to DE; for let HB, ME, be joined. Then because the Square of AH is equal to the Squares of AK and KH; and the Squares of AB, BK, are equal to the Square of AK; the Squares of AB, BK, KH, will be equal to the Square of AH; but the Square of BH is equal to the Squares of BK, KH; for the Angle HKB is a Right Angle, because HK is perpendicular to the Plane paffing through BA C. Therefore the Square of A H is equal to the Squares of AB, BH. Wherefore the Angle ABH is † a Right Angle. For the fame Reafon, the Angle DEM is alfo a Right Angle. And the Angle BAH is equal to the Angle EDM, for fo it is put; and AH is equal to DM. Therefore AB ist alfo equal to DE. And fo fince AC is equal to DF, and AB to DE, the two Sides CA, AB, fhall be equal to

*.26. 1.

48. 1.

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