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and 12. S.

CI. 6.

given similar to CD; therefore AB is similar to OK b. Then, Book XI. because AB is similar to CD, and the base AG to CN, as AE is to CF, so is EG to FN; and so 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 so is the parallelogram AG to GK; but, as GE is to EL, so is the parallelogram GK to KL “; and, as HE is to EM, so is the parallelogram PE to KM“; therefore, as AG is to GK, so is GK to KL; and fo is PE to KM; but, as AG is to GK, so is the solid AB to the solid EX d, for the d 25. plain GH is parallel to the opposite plains; and, as GK is to KL, so is the solid EX to PL d; and, as PE is to KM, so 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. II. is to EX, fo is the parallelogram AG to GK d; and so 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 solid KO is equal and similar to the folid CD, and the right line EK equal to CF : Therefore, the solid 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, so is a solid parallelopipedon described on the first, to a similar one described on the second.

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PRO P. XXXIV. THE O R.

T!

HE bases and altitudes of equal folid parallelopipedons are

reciprocally proportional ; and parallelopipedons, whose bases and altitudes are reciprocally proportional, arc equal.

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

First, let the insistent 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 base NP, so is CM to AG. For, if the base EH is equal to the base NP; then the altitudes CM, AG, are equal. For, if the bases EH, NP, are equal, but the altitudes not equal, then the solids are not equal“; but the solid 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 base NP, fo is CM to AGb. Now, let the bases be unequal, and let EH be the 32. greater, then the altitude CM will be greater than the altitude

AG;

C 7. 5.

d 32

32.

Book XI :G. for the solids are equa

Make CT equal to AG; and up. on PN compleat the solid cv, whose altitude is CT. Then, because the folids AB, CD, are equal, and CV is some other folid, AB is to CV as CD is to CV¢; but AB is to CV as the base

EH is to the base NP d; and, as the solid CD is to CV, so is the e I. 6. and base MP to TP, and so 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 bases and altitudes are reciprocally proportional, then AB is equal to CD. For, a gain, let the insistent right lines be at right angles to the bases; then, if the bases are equal, because the altitudes are as their bases, the altitudes are equal; therefore the solids AB, CD, are equal. But, if the bases are not equal, let EH be greater than NP, then the altitude of the solid CD is greater than the altitude of the solid AB ; that is, CM is greater than AG. Put CT equal to AG, and compleat the folid CV; then, because

the base EH is to the base NP, as MC is to AG, and CT is eI g. 5. qual to G; and, as FH is to NP, fo is MC to Crf; but, as

MC is to CT, fo is MP to PT e; and, as EH is to NP; fo is
AB to CV d; but, as AB is to CV, so is CD to CV; therefore
AB is equal to CD f.

Secondly, If the insistent right lines FE, BL, GA, KH, XN, DO, MC, RP, are not at right angles to the bases; from the points F, G, B, K, X, M, D, R, let be drawn perpendiculars meeting the plain of the bases EH, NP, in the points S, T, V, Y, Q, Z, a, f, and compleat the solids FV, Xa. Then, if the folids be equal, their bafes, and altitudes are reciprocally proportional, viz as EH is to NP, so 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 solid CD to the & 29. of 3o. folid DZ 8; therefore the solid BT is equal to the solid DZ; but

the bases and altitudes of equal solids, whose insistent right lines are at right angles to their bases, are proved to be reciprocally proportional. Therefore, as the bale KF is to the base RX, só is the altitude of the solid DZ to the altitude of the solid BT; but the folids DZ, DC, have the same altitude; and the folids BT, BA, have the same altitude ; therefore, the base EH is to the base NP, as the altitude of the solid DC is to the altitude of the folid AB.

Again, let the bases and altitudes of the solid parallelopipedons bé reciprocally proportional, viz, as LH is to NP, so let the altitude of CD be to the altitude of AB ; then the folids AB, CD, are equal. For, 'the same construction remaining, as the bafé EH is to the base NP, fo is the altitude of CD to the altitude of AB, but the altitudes of the folids AB, BT, are the same, and likewise of the solids CD, DZ; 'therefere, as the

base FK is to the base XR, so is the altitude of the solid 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 these folid parallelopipedons whose insistent right lines are at right angles to their bafes, and their bases and altitudes are reciprocally proportional, are equal to each other ; but the solid BT is equal to BA, for they stand upon the same base FK, and have the same altitude. For the same reason, the solid DZ is equal to the solid DC; therefore the solid AB is equal to the solid CD: Wherefore, folid parallelopipedons, whose bases and altitudes are reciprocally proportional, are equal. If the bafes are not equal, and the insistent right lines not at right angles to the bases, the bases and altitudes may be proved reciprocally proportional, in the same manner as when the infiftent right lines are at right angles to their bases; and in the same manner, if the bases and altitudes are reciprocally proportional, the folids are equal. Wherefore, &c.

PRO P. XXXV. THE O R.

If there be two plain angles equal, and from their vertices two

right lines be elevated 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 palling through the given right lines ; then these elevated lines shall 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 said 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 paffir.g thro' BAC; therefore HK is likewise perpendicular to the same 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 square of HA is equal to the squares of HK, KA, and the square of b 47. 1. KA equal to the squares of KC, CA b; therefore the square of AH is equal to the squares of HK, KC, CA; but the squares of HK, KC, are equal to the square of HC; for the angle HKC is a right one: Therefore the square of HA is equal to the squares of HC, CA; therefore the angles HCA, HBA, are right

ones.

e 26. I.

Book XI. ones. For the same reason, the angles DFM, DEM, are

right angles : Therefore the angle ACH is equal to DFM ; but C 48. J. d hyp.

the angle HAC & 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 side of the other; therefore, the re-
maining fides MF, FD, are equal to HC, CA, each to each
In the same manner, AB is proved equal to DE, and BH to
EM: Therefore, the two sides CA, AB, are equal to the two
fides FD, DE, and the angle BAC equal to FDE ; therefore the
base 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 same 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 sides DF, FN, and they
contain right angles; therefore the base AK is equal to DN.
And, since AH is equal to DM, their squares are equal; but
the squares of AK, KH, are equal to the square of AH, and the
squares of DN, NM, equal to the square of DM; for the angles
AKH, DNM, are right angles; therefore the squares of ĂK,
KH, are equal to the squares of DN, NM, and the squares of
AK, DN, equal ; therefore the squares of KH, MN, are equal ;

that is, the right line HK equal to MN; therefore the angle I S. So

HAK is equal to MDNf, that is, GAL equal to MDN:
Which was to be demonstrated.

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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IF three right lines A, B, C, are

proportional, the folid parallelo, pipedon made of them is equal to the solid parallelopipedon made of the middle line, it being equiangular to the former parallelopipedon.

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

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

a 261 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, so is GE to LX; therefore the sides about the equal angles GEF, MLX are reciprocally proportional; and the parallelograms MX, GF, equal b; and since the two plain angles b 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 passing through XLM, GEF“; therefore the folids LH, EK, have the same altitude ; but their bases MX, GF, are equal; therefore the solid HL is equal to the solid EK "; but HL is made d 31. of the three right lines A, B, C, and KE of the right line B. Wherefore, &c.

PRO P. XXXVII. THEO R.

IF four right lines are proportional, the similar solid parallelopi,

pedons described from them sball be proportional ; and if the similar solid parallelopipedons be proportional, then the right lines they are described

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 similar parallelopipedons described from them; then KA is to LC as MĚ is to NG.

For, because the folid parallelopipedon KA is similar to LC, KA is to LC in the triplicate proportion of AB to CD". Fora 33. the same reason, the solid ME is to NG in the triplicate proportion of EF to GH; but AB is to CD as EF to GH ; there-bun

. s. fore AK is to LC as ME to NG 6. 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 GH6. Wherefore, &c.

PRO P.

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