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and 12. 5.

C I. 6.

given fimilar to CD; therefore AB is fimilar to OK. Then, Book XI. because AB is fimilar to CD, and the base AG to CN, as AE m is to CF, fo is EG to FN; and fo is EH to FR ; and, because b 21. 6. FC is equal to EK, and FN to EL, and FR to EM, as AE is to EK, fo is EG to EL; and fo is the parallelogram AG to GK; but, as GE is to EL, fo is the parallelogram GK to KL ; and, as HE is to EM, fo is the parallelogram PE to KM ; therefore, as AG is to GK, fo is GK to KL; and fo is PE to KM; but, as AG is to GK, fo is the folid AB to the folid EX d; for the d 25. plain GH is parallel to the oppofite plains; and, as GK is to KL, fo is the folid EX to PL; and, as PE is to KM, fo is the folid PL to the folid KO. Then, because the four folids AB, EX, P, KO, are proportionals, AB has to KO a triplicate proportion of what AB has to EX ; but, as AB e def. 11. is to EX, fo is the parallelogram AG to GKd; and fo is the right line AE to the right line EK; therefore the folid AB has to the folid KO a triplicate ratio of what AE has to EK; but the folid KO is equal and fimilar to the folid CD; and the right line EK equal to CF: Therefore, the folid AB has to the folid CD a triplicate ratio of what the homologous fide AE has to the homologous fide CF. Wherefore, &c.

COR. Hence, if four right lines be proportional, as the first is to the fourth, fo is a folid parallelopipedon defcribed on the first, to a fimilar one described on the fecond.

PRO P. XXXIV. THEOR.

THE bafes and altitudes of equal folid parallelopipedons are reciprocally proportional; and parallelopipedons, whofe bafes and altitudes are reciprocally proportional, are equal.

Let AB, CD, be equal folid parallelopipedons; then the bafe EH is to the base NP, as the altitude of the folid CD is to the altitude of the folid AB.

First, let the infiftent right lines AG, EF, LB. HK, CM, NX, OD, PR, be at right angles to their bases; then, as the bafe EH is to the bafe NP, fo is CM to AG. For, if the base EH is equal to the bafe NP; then the altitudes CM, AG, are equal. For, if the bases EH, NP, are equal, but the altitudes not equal, then the folids are not equal; but the folid AB is a 31, put equal to the folid CD; therefore the altitudes CM, AG, are equal: Therefore, as the base EH is to the bafe NP, fo is CM to AG. Now, let the bafes be unequal, and let EH be the 32. greater, then the altitude CM will be greater than the altitude

AG;

Book XIG for the folids are equa

C 7. 5.

32,

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Make CT equal to AG; and upon PN compleat the folid CV, whofe altitude is CT. Then, becaufe the folids AB, CD, are equal, and CV is fome other folid, AB is to CV as CD is to CV; but AB is to CV as the bafe d 32. EH is to the bafe NP 4; and, as the folid CD is to CV, fo is the e 1. 6. and bafe MP to TP, and fo is MC to TC; therefore the base EH is to NP as MC is to CT; but CT is equal to AG; therefore EH is to NP as MC is to AG. And, if the bafes and altitudes are reciprocally proportional, then AB is equal to CD. For, a gain, let the infiftent right lines be at right angles to the bases; then, if the bafes are equal, because the altitudes are as their bafes, the altitudes are equal; therefore the folids AB, CD2, are equal. But, if the bafes are not equal, let EH be greater than NP, then the altitude of the folid CD is greater than the altitude of the folid AB; that is, CM is greater than AG. Put CT equal to AG, and compleat the folid CV; then, because the bafe EH is to the bafe NP, as MC is to AG; and CT is equal to AG; and, as EH is to NP, fo is MC to Cff; but, as MC is to CT, fo is MP to PT ; and, as EH is to NP, fo is AB to CV d; but, as AB is to CV, fo is CD to CV; therefore AB is equal to CD f.

f 9. 5.

€ 29. or 30.

Secondly, If the infiftent right lines FE, BL, GA, KH, XN, DO, MC, RP, are not at right angles to the bafes; from the points F, G, B, K, X, M, D, R, let be drawn perpendiculars meeting the plain of the bafes EH, NP, in the points S, T, V, Y, Q, Z, a, f, and compleat the folids FV, Xa. Then, if the folids be equal, their bafes and altitudes are reciprocally proportional, viz as EH is to NP, fo is the altitude of CD to the altitude of A* f. For, because the folids AB, CD, are equal, and the folid AB is equal to the folid BT, and the folid CD to the folid DZ ; therefore the folid BT is equal to the folid DZ; but the bafes and altitudes of equal folids, whofe infiftent right lines are at right angles to their bafes, are proved to be reciprocally proportional. Therefore, as the bafe KF is to the base RX, fo is the altitude of the folid DZ to the altitude of the folid BT; but the folids DZ, DC, have the fame altitude; and the folids BT, BA, have the fame altitude; therefore, the bafe EH is to the base NP, as the altitude of the folid DC is to the altitude of the folid AB.

Again, let the bafes and altitudes of the folid parallelopipedons be reciprocally proportional, viz. as EH is to NP, fo let the altitude of CD be to the altitude of AB; then the folids AB, CD, are equal. For, the fame construction remaining, as the bafe EH is to the bafe NP, fo is the altitude of CD to the altitude of AB; but the altitudes of the folids AB, BT, are the fame, and likewife of the folids CD, DZ; therefore, as the

bafe FK is to the bafe XR, fo is the altitude of the folid DZ to Book XI. the altitude of the folid BT; therefore the base and altitudes of the folid parallelopipedons BT, DZ, are reciprocally proportional: But thefe folid parallelopipedons whofe infiftent right lines are at right angles to their bafes, and their bafes and altitudes are reciprocally proportional, are equal to each other; but the folid BT is equal to BA, for they ftand upon the fame base FK, and have the fame altitude. For the fame reason, the folid DZ is equal to the folid DC; therefore the folid AB is equal to the folid CD: Wherefore, folid parallelopipedons, whofe bafes and altitudes are reciprocally proportional, are equal. If the bafes are not equal, and the infiftent right lines not at right angles to the bafes, the bafes and altitudes may be proved reciprocally proportional, in the fame manner as when the infiftent right lines are at right angles to their bafes; and in the fame manner, if the bafes and altitudes are reciprocally proportional, the folids are equal. Wherefore, &c,

IF

PRO P. XXXV. THE OR.

F there be two plain angles equal, and from their vertices two right lines be levated above the plains in which the angles are, making equal angles with the lines containing the given angles; and if, in the elevated lines, any points be taken, from which perpendiculars are let fall to the plains paffing through the given right lines; then thefe elevated lines fhall be equally inclined to the given plain.

Let BAC, EDF, be two right lined plain angles, from whose vertices A, D, let two right lines AG, DM, be elevated above the plains of the faid angles, making the angles MDE, MDF, equal to the angles GAB, GAC, each to each; then the angle MDN will be equal to GAL.

For, if AG is not equal to DM, take AH equal to DM; and from H draw HK parallel to GL; but GL is perpendicular to the plain paffing thro' BAC; therefore HK is likewife perpendicular to the fame plain. From the points K, N, draw the right lines a 7. KB, KC, NE, NF, perpendicular to the right lines AB, AC, DE, DF. Join HC, CB, MF, EF, HB, EM; then the fquare of HA is equal to the fquares of HK, KA, and the fquare of b 47. x. KA equal to the fquares of KC, CA; therefore the fquare of AH is equal to the fquares of HK, KC, CA; but the fquares of HK, KC, are equal to the fquare of HC; for the angle HKC is a right one: Therefore the fquare of HA is equal to the fquares of HC, CA; therefore the angles HCA, HBA, are right

ones.

€ 48. 1.

d hyp.

€ 26. I.

Book XI. ones. For the fame reason, the angles DFM, DEM, are right angles: Therefore the angle ACH is equal to DFM; but the angle HAC d is equal to MDF; therefore the two triangles HAC, MDF, have two angles, HAC, HCA, in the one, equal to MFD, MDF, of the other, each to each, and HA, a fide of the one, equal to MD, a fide of the other; therefore, the remaining fides MF, FD, are equal to HC, CA, each to each *. In the fame manner, AB is proved equal to DE, and BH to EM: Therefore, the two fides CA, AB, are equal to the two fides FD, DE, and the angle BAC equal to FDE; therefore the bafe BC is equal to EF, the angle ACB to DFE, and ABC to DEF, But the angles ACK, DFN, are equal; for each is a right one; therefore the remaining angle BCK is equal to EFN. For the fame reason, the angle CBK is equal to FEN. And, because the two triangles BCK, EFN, have the two angles CBK, BCK, in the one, equal to the two angles FEN, EFN, each to each, and a fide BC equal to EF; therefore the fide FN is equal to CK. But AC is equal to DF; therefore the two fides AC, CK, are equal to the two fides DF, FN, and they contain right angles; therefore the base AK is equal to DN. And, fince AH is equal to DM, their fquares are equal; but the fquares of AK, KH, are equal to the fquare of AH, and the fquares of DN, NM, equal to the square of DM; for the angles AKH, DNM, are right angles; therefore the squares of AK, KH, are equal to the fquares of DN, NM, and the fquares of AK, DN, equal; therefore the fquares of KH, MN, are equal; that is, the right line HK equal to MN; therefore the angle HAK is equal to MDNf, that is, GAL equal to MDN: Which was to be demonftrated.

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COR. Hence, if two right lined plain angles be equal, from whose point equal right lines are elevated on the plain of the angles containing equal angles with the given lines, each to each; perpendiculars drawn from the extreme points of these elevated lines to the plain of the given angles, are equal to one another.

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F three right lines A, B, C, are proportional, the folid parallelopipedon made of them is equal to the folid parallelopipedon made of the middle line, it being equiangular to the former parallelopipedon.

Let the three proportional right lines be A, B, C, viz. A to Book XI. Bas B is to C; then the folid made with A, B, C, is equal to the equiangular folid made from B.

Let E be a folid angle contained under the plain angles DEF, GEF, GED; make ĜE, EF, DE, each equal to B ; and compleat the folid parallelopipedon EK. Again, put LM equal to A; and at the point L, with the right line LM, make a folid angle contained under the three plain angles NLX, XLM, MLN, equal to the folid angle Ea.

a 26

Then, because LM, LN, EF, EG, or ED, and LX, are equal to the three proportional lines A, B, C; therefore, as LM is to EF, fo is GE to LX; therefore the fides about the equal angles GEF, MLX are reciprocally proportional; and the parallelograms MX, GF, equal; and fince the two plain angles ь 14.6. GEF, XLM, are equal, and from the points N, D, are drawn the equal perpendiculars NL, DE, at right angles to the plain paffing through XLM, GEF; therefore the folids LH, EK, have the fame altitude; but their bases MX, GF, are equal; therefore the folid HL is equal to the folid EK4; but HL is made d 31. of the three right lines A, B, C, and KE of the right line B. Wherefore, &c.

PROP. XXXVII. THEOR.

IF four right lines are proportional, the fimilar folid parallelopi pedons defcribed from them fhall be proportional; and if the fimilar folid parallelopipedons be proportional, then the right lines they are defcribed from shall be proportional.

Let the four right lines AB, CD, EF, GH, be proportional, viz. AB to CD, as EF is to GH; and let KA, LC, ME, NG, be the fimilar parallelopipedons defcribed from them; then KA is to LC as ME is to NG.

For, because the folid parallelopipedon KA is fimilar to LC, KA is to LC in the triplicate proportion of AB to CD1. For1 33. the fame reason, the folid ME is to NG in the triplicate proportion of EF to GH; but AB is to CD as EF to GH; there- . s. fore AK is to LC as ME to NG . And, if AK be to LC as ME to NG, then AB is to CD as EF is to GH. For, because AK has to LC a triplicate proportion of AB to CD, and ME to NG, a triplicate proportion of EF to GH; therefore AB is to CD as EF to GH. Wherefore, &c.

PRO P.

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