Points S, T, Y, V, Q, Z, o, §, and compleat the Solids FV, X2. 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 Base 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 *30ofthis. CD, and the Solid AB is equal to the Solid BT; for they stand upon the fame Base, 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 ftand upon the fame Bafe, X R 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 Bases and Altitudes of thofe equal Solids, whofe Altitudes are at Right +From Angles to their Bafes, are‡ reciprocally proportional. what has Therefore as the Bafe FK is to the Bafe XR, fo is been before the Altitude of the Solid DZ, to the Altitude of the proved. Solid BT; but the Bafe F K is equal to the Base EH, and the Bafe XR to the Bafe 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 BT; but the Solids DZ, DC, have the fame Altitude, and fo have the Solids BI, 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.

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 AB. I fay, the Solid AB is equal to the Solid CD.

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

Altitude

Altitude of the Solid DZ is to the Altitude of the Solid BT; wherefore the Bases 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 Bafe 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 Base XR, have the fame Altitude, and their infiftent Lines are not in the fame Right Lines. Therefore the Solid AB is equal to the Solid CD; which was to be demonstrated.

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 correSpondent 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.

LET BAC, EDF, be two equal Right-lin❜d An

gles; and from A, D, the Vertices of thofe Angles, let two Right Lines A G, D M, 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 MDE equal to the Angle G A B, and the Angle MDF to the Angle GAC; and take any Points G and M in the Right Lines A G, D M, from which let GL, MN, be drawn perpendicular to the Planes paffing thro BAC, EDF, meeting the fame in the Points L, N, and

Q 3

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 parallerto GL; but GL is perpendicular to the Plane paffing thro' BAC. Therefore HK fhall *8of 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, K A, 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 also 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 26.1. 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 A B is equal to DE; for let HB, ME, be joined. Then because the Square of A H is equal to the Squares of AK and K H; 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 H K is perpendicular to the Plane paffing thro' BA C. Therefore the Square of AH is equal to the Squares of AB, BH. Wherefore the Angle ABH is a Right Angle. For the fame Rcafon, the Angle DE M is alfo a Right Angle. And the Angle B A H is equal to the Angle EDM, for fo it is put; and AH is equal to DM. Therefore AB is alfo equal to DE. And fo fince AC is equal to DF, and AB to DE, the two Sides C A, AB, fhall be equal to

+48. 1.

*

*

the

the two Sides FD, DE; but the Angle BAC is equal to the Angle FDE. Therefore the Bafe BC is equal to the Base 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 EFN. For the fame Reafon, 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 shall have the other Sides equal to the other Sides. Therefore CK is equal to FN, but A C is equal to DF. Therefore the two Sides AC, 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 DN. And fince AH is equal to DM, the Square of AH fhall be equal to the Square of DM; but the Squares of AK, KH, are equal to the Square of AH; 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 A K, K H, 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 Base NM, the Angle HAK fhall 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-lin❜d plane Angles equal, from whofe Points equal Right Lines be elevated on the Planes of the Angles, containing equal Angles with the Lines firft 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. PRO

Q 4

PROPOSITION XXXVI.

THEOREM.

If three Right Lines be proportional, the Solid 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.

LET 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.

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 compleat the folid Parallelepipedon EK. Again, put LM equal to A, 26 of this. and at the Point L, at the Right Line LM, make * a 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 because A is to B, as B is to C, and A is equal to LM, and B to LN, EF, EG, or ED, and C to LX; it fhall 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 14. 6. MX is equal to the Parallelogram GF. And fince the two plane Angles GEF, XL M, are equal, and the Right Lines LN, ED, being equal are erected at the angular Points containing equal Angles with the Lines first given, each to each; the Perpendiculars drawn from the Points N, D, to the Planes drawn thro' XLM, GEF, are equal one to another. Thèrefore 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

Cor. 35

of this.

*