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BOOK XI. .
PROP. XXX. THEOR.
OLID parallelopipeds upon the fame base, and of
the same altitude, the insisting straight lines * of which are not terminated in the same straight lines, are equal to one another.
Let the parallelopipeds CM, CN be upon the same base AB, and of the same altitude; but their infifting straight lines AF, AG, LM, LN, CD, CE, BH, BK not terminated in the same straight lines; the solids CM, CN are equal.
Produce FD, MH, and NG, KE, and let them meet one another in the points O, P, Q, R; and join AO, LP, BO, CR: and because the plane LBHM is parallel to the opposite plane ACDF, and that the plane LBHM is that in which are the parallels LB, MPHQ, in which also is the figure BLPQ; and the plane ACDF is that in which are the parallels AC, FODR, in which also is the figure CAOR; therefore the figures BLPQ: CAOR are in parallel planes: In like manner, it may be shewn, that the figures ALPO, CBQR are in parallel planes : and the planes ACBL, OROP are parallel ; therefore the folid CP is a
parallelopiped : But the solid CM, of which the base is ACBL, a 29. 11. to which FDHM is the opposite parallelogram is equal a to the folid CP, of which the base is the parallelogram ACBL, to
* The insisting straight lines are the sides of the parallelograms betwixt the base and the opposite plane.
which OROP is the one opposite, because they are upon the Book XI. fame base, and their infisting straight lines AF, AO, CD, CR; LM, LP, BH, BQ are in the same straight lines FR, MQ: and the solid CP is equal a to the folid CN; for they are upon the a 29. II. fame base ACBL, and their infifting straight lines AO, AG, LP, LN; CR, CE, BQ, BK are in the same straight lines ON, RK; therefore the solid CM is equal to the solid CN. Wherefore, solid parallelopipeds, &c. Q. E. D.
PROP. XXXI. THEOR.
OLID parallelopipeds which are upon equal bases,
and of the same altitude, are equal. Let the folid parallelopipeds AE, ÇF be upon equal bases AB, CD, and be of the fame altitude; the folid AE is equal to the folid CF.
First, Let the insisting straight lines be at right angles to the bases AB, CD, and let the bases be equiangular, having the angle ALB equal to CLD, and let them be placed in the same plane, and so as that CL, LB be in a straight line; therefore AL, LD are in a straight line a ; and the straight line LM, which is
Cor. at right angles to the plane in which the bases are in the point is. I. to the
b 13. II. two folids AE, CF; let
R the other insisting lines of the folids be AG, HK,
E BE; DF, OP, CN. Produce OD, HB, and let them meet in Q and complete the parallelo
B piped LR, of which the base is the parallelogram
Н LQ, and LM is one of its infifting lines : Therefore, because the parallelogram AB is equal to CD, as AB to LQ, so is CD to LQ : and because the C7. 5. parallelopiped AR is cut by the plane LMEB, which is parallel to the opposite planes AK, DR; as the base AB to the base LQ, so is d the folid AE to the solid LR: and because the parallelo- d 25. 11. piped CR is cut by the plane LMFD parallel to the opposite planes CP, BR; as the base CD to the base LQ, fo is d the folid CF to the solid LR: But as AB to LQ: so is CD to LQ, as before was proved ; therefore as the folid AE to the folid LR, so is the solid CF to the solid LR; and therefore the folid AE e 11. 5 is equal to the solid CF.
f 9.5 Next,
Book XI. Next, Let the insisting lines be at right angles to the bases,
but the bases AB, CD not be equiangular. At the point L in 8 23. 1. the straight line LB, makes the angle BLS equal to CLD, and
let LS meet IIA in S, and complete the parallelogram BS, and the parallelopiped SE, of which the base is SB, and one of the
infifting lines is LM: Then, the solid SE is equal to the solid h 29. 11. AE 5, because they are upon the same base LE, and of the
fame altitude, and their inlifting lines LS, LA, BT, BH; MV,
MG, EX, EK are in the same straight lines SH, VK: and bek 35. 1. cause the parallelogram AB is equal to SB, for they are upon
the same base LB, and between the same parallels LB, AT; and that the base AB is equal to the base CD, therefore SB is equal to CD, and the angle SLB is equal to the angle CLD: there. fore, by the first case, the solid SE is equal to the solid CF: but the folid SE is equal to the solid AE, as was demonstrated ; therefore the folid AE is equal to the solid CF.
But, if the insisting lines AG, HK, BE, LM; CN, RS, DF, OP be not at right angles to the bases AB, CD: From the
111.11. points G, K, E, M; N, S, F, P draw ? GQ, KT, EV, MX;
NY, SZ, FI, PU perpendiculars to the plane in which are the bases AB, CD, and let them meet it in the points Q, T, V, X; Y, Z, I, U, and join QT, TV, VX, XQ; YZ, ZI, IU, UY:
Then, because QG, KT are at right angles to the same plane, m 6. 11. they are parallels : and MG, EK are parallels; therefore the n 15. 11. planes MO, ET passing through them are parallel - to one another: For the same reason, the planes MV, GT are parallel;
therefore the solid QE is a parallelopiped. In like manner, it Book XI. may be proved, that the folid YF is a parallelopiped : But, from what has been demonstrated, the solid EQ
is equal to the solid FY, because they are upon equal bases MK, PS, and of the same altitude, and have their insisting lines at right angles to the bases: and the folid EQ is equal o to the folid AE ; and the o 29. or solid FY to the solid CF, because they are upon the same bases, 30.11. and of the same altitude: Therefore the folid AE is equal to the solid CF. Wherefore solid parallelopipeds, &c. Q. È. D.
PROP. XXXII, THEOR.
tude, are to one another as their bases.
Let AB, CD be solid parallelopipeds of the same altitude; they are to one another as their bases; that is, as the base AE to the base CF, so is the solid AB to the folid CD.
Produce CG to H; and to FG apply the parallelogram FH equal to AE, having FGH for one of its angles; and complete a Cor.45.1.
the parallelopiped GK upon the base FA, and having FD for one of its insisting lines; therefore the solids CD, GK are of the same altitude; therefore the fold AB is equal to the folid GK, b 31.11. because they are upon equal bases AE, FH, and are of the same altitude: and because the solid CK is cut by the plane DG, which is parallel to its oppofite planes, the base HF is to the base FC, as the folid HD to the solid DC; But the base HF C 25. 11, is equal to the base AE, and the folid GK to the solid AB; therefore, as the base AE to the base CF, so is the solid AB to the folid CD. Wherefore, &c. Q. E. D.
Cor. 1. From this it is manifest, that prisms upon triangular bases, of the same altitude, are to one another as their bases.
Let the prisms AEM-NBO and CFG-PDQ have the same altitude; and complete the parallelograms AE, CF, and the parallelopipeds AB, CD, of which the insisting lines are MO, GQ:
Book XI. and because the parallelopipeds have the same altitude, AB is to
CD, as the base AE to CF; wherefore the prisms, which are d 28. 11. their halves 4, are to one another, as AE to CF; that is, as the triangle AEM to CFG.
Also, any prisms of the same altitude are to one another as their bases: For prisms upon any other figures may be divided into prisms having triangular bases : And each tri
angle is to each triangle, as the prism upon the first to the prism eCor.24.5. upon the other ; therefore the whole prism is to the whole as
the base of the first to the base of the other.
PROP. XXXIII. THEOR.
IMILAR folid parallelopipeds are one to another in
the triplicate ratio of their homologous fides. Let AB, CD be fimilar parallelopipeds, and the fide AE ho. mologous to the side CE: The folid AB has to the solid CD, the triplicate ratio of that which AE has to CE.
Let the bases AG, CN be in the same plane, and the solids AB, CD on different sides of it, and let AE, EC be in a straight
line; therefore, because the angles AEG, CEN are equal, GE, 2 3. Cor. EN are in a traight line a : And if the plane ED be produced 15. 1.
beyond GN, it shall coincide with the plane GH, because these
G the parallelogram GC and the pa. rallelopipeds EX, NP upon the
С bases GC, CN, so that EH be an
R because the solids AB, CD are fi
D milar, as AE to EG, fo is CE to EN; and, by alternation, as AE to EC, so is GE to EN: For
the same reason, as AE to EC, so is HE to ER: But as AE to b 1.6. EC, so b is the parallelogram AG to GC, and the folid AB to
the solid EX, because they have the same altitude; and as GE C32• 11.
to EN, fob is the base GC to the base CN, and the solid EX to the solid NPS: and as HE to ER, so b is the base HC to the base CR, and the solid NP to the folid CD C: But as AE to EC,
so is GE to EN, and HE to ER; therefore, as the folid AB to d is. 5. the folid EX, so dis EX to NP, and NP to CD: But if four