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Book XI. Produce MA, NA, OA to P, Q, R, so that AP may be

equal to DL, AQ to DK, and AR to DH; and complete the solid parallelepiped AX contained by the parallelograms

AS, AT,' AV similar and equal to CH, CK, CL, each to a C. 11. each. Therefore the solid AX is equal a to the solid CD.

Complete likewise the solid AY; the base of which is AS,

and of which AO is one of its insisting straight lines. Take e 12. 6. any straight line a, and as MA to AP, so make a to b; and

as NA to AQ, so make b to c; and as AO to AR, so make c to d: Then, because the parallelogram AE is equiangular to AS, AE is to AS, as the straight line a to c, as is demonstrated in the 23d Prop. Book VI. and the solids AB, AY,

being betwixt the parallel planes BOY, EAS, are of the same b 32. 11. altitude. Therefore the solid AB is to the solid AY, as b the

base AE to the base AS; that is, as the straight line a is to c.

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c 25. 11. And the solid AY is to the solid AX, as the base OQ is to fl. 6. the base QR; that is, as the straight line OA to AR'; that

is, as the straight line c to the straight line d. And because the solid AB is to the solid AY, as a is to c, and the solid AY to the solid AX, as c is to d; ex æquali, the solid AB is to the solid AX, or CD which is equal to it, as the straight

line a is to d. But the ratio of a to d is said to be compoundd def. A. 5 ed d of the ratios of a to b, b to c, and c to d, which are the

same with the ratios of the sides MA to AP, NA to AQ, and OA to AR, each to each. And the sides AP, AQ, AR are equal to the sides DL, DK, DH, each to each. Therefore the solid AB has to the solid CD the ratio which is the same with that which is compounded of the ratios of the sides AM to DL, AN to DK, and AO to DH. Q. E. D.

Book XI.

PROP. XXXIV. THEOR.

The bases and altitudes of equal solid parallele. See N. pipeds, are reciprocally proportional ; and if the bases and altitudes are reciprocally proportional, the solid parallelepipeds are equal.

Let AB, CD be equal solid parallelepipeds; their bases are reciprocally proportional to their altitudes: that is, as the base EH is to the base NP, so is the altitude of the solid CD to the altitude of the solid AB.

First, Let the insisting straight lines AG, EF, LB, HK:
CM, NX, OD, PR be at right angles to the bases. As the base
EH to the base NP, so
is CM to AG. If the

K B R D
base EH be equal to the
base NP, then because G E M

X the solid AB is likewise equal to the solid CD, CM will be equal to H

LP 10 AG. Because if the bases EH, NP be

А

C N equal, but the altitudes AG, CM be not equal, neither will the solid AB be equal to the solid CD. But the solids are equal by the hypothesis. Therefore the altitude CM is not unequal to the altitude AG; that is, they are equal. Wherefore, as the base EH to the base NP, so is CM to AG.

Next, Let the bases EH, NP not be equal, but EH greater than the other: Since then the solid AB is equal to the solid CD, CM is therefore greater than AG: For

R D if it be not, neither also in this case would K B

X the solids AB, CD be equal, which, by the

G F hypothesis, are equal. Make then CT equal to AG, and complete H

P

0 the solid parallelepiped CV, of which the base Α. E c N is NP, and altitude ĆT. Because the solid AB is equal to the solid CD, therefore the

Book XI. solid AB is to the solid CV, as a the solid CD to the solid CV.

But as the solid AB to the solid CV, sob is the base EH to a 7. 5

the base NP; for the solids AB, CV, are of the same altitude; b 32. 11.

and as the solid CD to CV, so c is the base MP to the base c 25. 11. d l. 6. PT; and so d is the straight line MC to CT; and CT is equal

to AG. Therefore, as the base EH to the base NP, so is MC to AG. Wherefore the bases of the solid parallelepipeds AB, CD are reciprocally proportional to their altitudes.

Let now the bases of the solid parallelepipeds AB, CD, be
reciprocally proportional to their altitudes : viz. as the base EH
to the base NP, so the
altitude of the solid CD

K B R D
to the altitude of the so-
lid AB; the solid AB is G

M

X
equal to the solid CD.
Let the insisting lines
be, as before, at right H L P

0
angles to the bases.
Then, if the base EH

A E с N
be equal to the base NP,
since EH is to NP, as

the altitude of the solid CD is to the altitude of the solid AB, e A. 5. therefore the altitude of CD is equal e to the altitude of AB.

But solid parallelepipeds upon equal bases, and of the same f31. 11. altitude, are equal to one another; therefore the solid AB

is equal to the solid CD.

But let the bases EH, NP be unequal, and let EH be the
greater of the two. Therefore, since as the base EH to the base
NP, so is CM the al-
titude of the solid CD to

R D
AG the altitude of AB,
CM is greater than

K B

M X
AG. Again, take CT
equal to AG, and com-

F
plete, as before, the so-
lid CV. And, because
the base EH is to the H

L

P
base NP, as CM to
AG, and AG is equal to А. E CN
CT, therefore the base
EH is to the base NP, as MC to CT. But as the base EH is to
NP, sob is the solid AB to the solid CV; for the solids AB,

CV are of the same altitude : and as MC to CT, so is the base c 25. 11. MP to the base PT, and the solid CD to the solid · CV; And therefore as the solid AB is to the solid CV, .so is the Book XI. solid CD to the solid CV, that is, each of the solids AB, CD has the same ratio to the solid CV; and therefore the solid AB is equal to the solid CD.

Second general case. Let the insisting straight lines FE, BL, GA, KH; XN, DO, MC, RP not be at right angles to the bases of the solids; and from the points F, B, K, G; X, D, R, M draw perpendiculars to the planes in which are the bases EH, NP meeting those planes in the points S, Y, V, T; R, I, U, Z; and complete the solids FV, XU, which are parallelepipeds, as was proved in the last part of Prop. 31. of this Book. In this case, likewise, if the solids AB, CD be equal, their bases are reciprocally proportional to their altitudes, viz. the base EH to the base NP, as the altitude of the solid CD to the altitude of the solid AB. Because the solid AB is equal to the solid CD, and also to the solid BT6, for

29. or they are upon the same base FK, and of the same altitude; 30. 11.

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and because the solid DC is equal to the solid DZ, being upon the same base XR, and of the same altitude, therefore the solid BT is equal to the solid DZ: But the bases are · reciprocally proportional to the altitudes of equal solid parallelepipeds, of which the insisting straight lines are at right angles to their bases, as before was proved: Therefore, as the base FK to the base XR, so is the altitude of the solid DZ to the altitude of the solid BT: And the base FK is equal to the base EH, and the base XR to the base NP: Wherefore, as the base EH to the base NP, so is the altitude of the solid DZ to the altitude of the solid BT: But the altitudes of the solids DZ, DC, as also of the solids BT, BA are the same. Therefore, as the base EH to the base NP, so is the altitude

Book XI. of the solid CD to the altitude of the solid AB; that is, the

bases of the solid parallelepipeds AB, CD are reciprocally proportional to their altitudes.

Next, Let the bases of the solids AB, CD be reciprocally proportional to their altitudes, viz. the base EH to the base NP, as the altitude of the solid CD to the altitude of the solid AB; the solid AB is equal to the solid CD: The same construction being made; because as the base EH to the base NP, so is the altitude of the solid CD to the altitude of the solid AB; and because the base EH is equal to the base FK; and NP to XR; therefore, the base FK is to the base XR, as the altitude of the solid CD to the altitude of AB:

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But the altitudes of the solids AB, BT are the same, as also of CD and DZ; therefore, as the base FK to the base XR, so is the altitude of the solid DZ to the altitude of the solid BT: Wherefore the bases of the solids BT, DZ are reciprocally proportional to their altitudes; and their insisting straight lines are at right angles to the bases; wherefore, as was be

fore proved, the solid BT is equal to the solid DZ: But BT 8 29. or 30. is equal to the solid BA; and DZ to the solid DC, because 11.

they are upon the same bases, and of the same altitude. Therefore the solid AB is equal to the solid CD. Q. E. D.

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