angles to the base CD, their intersection LM is at right angles to the plane CD (Prop. 34); and for the same reason, the intersection of the planes AM, BM is at right angles to the plane AB, that is to the plane CD; consequently, LM is also the intersection of the planes AM, BM (Prop. 8).

First, let the angle ALB be equal to the angle CLD (fig. 62); then AL, LD are in a straight line (I. 14). Produce OD, HB, and let them meet in Q; and complete the parallelepiped LR, of which the base is LQ.

Because the parallelogram AB is equal to the parallelogram CD, as the base AB is to the base LQ, so is the base CD to the base LQ (V. 7). And because the parallelepiped AR is cut by the plane LMEB, which is parallel to the opposite planes AK, DR; as the base AB is to the base LQ, so is the solid AE to the solid LR: for the same reason, because the parallelepiped CR is cut by the plane LMFD which is parallel to the opposite planes CP, BR; as the base CD is to the base LQ, so is the solid CF to the solid LR: but it has been shown that as the base AB is to the base LQ, so is the base CD to the base LQ; therefore as the solid AE is to the solid LR, so is the solid CF to the solid LR (V. 11): consequently the solid AE is equal to the solid CF (V. 9).

Next let the angles ALB, CLD of the equal bases AB, CD, be unequal (fig. 63). Produce HA, DL until they meet in S; from B draw BT parallel to DS; and let TB, OD produced meet in Q; and complete the parallelepipeds LX, LR : then the parallelepiped SE, of which the base is the parallelogram LE, and SX the plane opposite, is equal to the parallelepiped AE, of which the base is LE, and AK the plane opposite; for they are upon the same base, and the parallelograms have two of their sides opposite to the base in the same straight line VK (Prop. 55). And because the parallelogram SB is equal to the parallelogram AB (I. 35); and that the base AB is, by hypothesis, equal to the base CD; the base SB is equal to the base CD: but the angle SLB is equal to the angle CLD (I. 15); therefore by the former case the solid SE is equal to the solid CF and it has been shown that the solid SE is equal to the solid AE; therefore the solid AE is equal to the solid CF.

:

Case 2.-Let the planes AK, HE, BR, QG and CM, LF, DP, ON (fig. 64) of the parallelepipeds AE and CF, adjacent to their bases AB and CD, be not at right angles to them: in this case likewise the parallelepiped AE is equal to the parallelepiped CF.

Because parallelepipeds which are upon the same base, and have their planes opposite to the base in the same plane, are equal (Prop. 55), if two parallelepipeds be constituted on the bases AB and CD by planes at right angles to them, and have their planes opposite to the bases AB and CD in the same planes as GE and NF respectively, they will be equal to the solids AE and CF re

spectively, and, by the first case, they will be equal to one another; therefore the parallelepipeds AE and CF are also equal to one an

other.

Wherefore parallelepipeds which are upon equal bases, &c.: which was to be proved.

Cor. Any parallelepiped is equal to a rectangular parallelepiped, (that is, a parallelepiped every two adjacent planes of which are perpendicular to each other,) upon an equal base and of the same altitude.

PROP. LVII. THEOR.

Parallelepipeds which have the same altitude are to one another as their bases.

Let AB, CD (fig. 65) be parallelepipeds which have the same altitude: they are to one another as their bases; that is,

The solid AB is to the solid CD, as the base AE to the base CF. To the straight line FG apply the parallelogram FH equal to AE, and having the angle FGH equal to the angle LCG (I. 45. Cor.), so that CG, GH is a straight line; and complete the parallelepiped GK between the planes CFH, PDM, so that GK is of the same altitude as CD, and therefore as AB. Then the solid GK is equal to the solid AB, because they are upon equal bases and have the same altitude (Prop. 56). And because the parallelepiped CK is cut by the plane DG, parallel to its opposite planes, the solid GK is to the solid CD as the base FH to the base CF: but the solid GK is equal to the solid AB, and the base FH to the base AE; therefore the solid AB is to the solid CD as the base AE to the base CF. Wherefore parallelepipeds which have the same altitude, &c.: which was to be proved.

PROP. LVIII. THEOR.

Parallelepipeds are to one another in the ratio that is compounded of the ratios of their bases, and of their altitudes.

Let AF, GO (fig. 66) be two parallelepipeds, of which the bases are AC and GK, and the altitudes, the perpendiculars let fall on the planes of these bases from any point in the opposite planes EF and MO: the solid AF is to the solid GO in the ratio compounded of the ratios of the base AC to the base GK, and of the perpendicular on AC to the perpendicular on GK.

Case 1.-Let the solids AF, GO be right parallelepipeds, so that AE and GM, being at right angles to the bases AC and GK, are the altitudes of the solids. In GM take GQ equal to AE, and through Q let a plane pass parallel to GK, meeting the planes GN, HO, KP, LM in QR, RS, ST, TQ. GS is therefore a parallelepiped, and it has the same altitude as AF, viz. GQ or AE.

To GH apply the parallelogram HX equal to AC, so that GLX

is a straight line (I. 45. Cor.), and complete the parallelepiped GV, which is equal to the parallelepiped AF, because they are upon equal bases HX, AC, and of equal altitudes GQ, AE (Prop. 56). Take GY a fourth proportional to GQ or AE, GM, GL (VI.12), that is GQ or AE to GM as GL to GY.

Because GX is to GL as the parallelogram HX to the parallelogram GK (VI. 1), and that HX is equal to the base AC, the ratio GX to GL is the ratio of the base AC to the base GK; also GL to GY is the ratio of the altitude AE to the altitude GM. Now the ratio of GX to GY is the ratio compounded of the ratios of GX to GL and GL to GY (V. Def. 11. A.); consequently the ratio of GX to GY is the ratio compounded of the ratios of the base AC to the base GK, and of the altitude AE to the altitude GM. Also the solid GV is to the solid GS, as the base HX to the base GK (Prop. 57), that is as GX to GL; and since the solid GV is equal to the solid AF, the solid AF is to the solid GS as GX to GL. But the solid GS is to the solid GO as GQ to GM (Prop. 57), that is as GL to GY. Since then

and

the solid AF : the solid GS :: GX : GL the solid GS: the solid GO :: GL : GY ex æquali, the solid AF: the solid GO :: GX: GY (V. 22).

But it has been shown that GX has to GY the ratio compounded of the ratios of the base AC to the base GK, and of the altitude AE to the altitude GM; therefore the solid AF has to the solid GO, the ratio compounded of the ratios of the base AC to the base GK, and of the altitude AE to the altitude GM.

Case 2.-When the solids are not right parallelepipeds.

Let the parallelograms AC and GK be the bases, and AE, GM the altitudes, of two oblique parallelepipeds W, Z. Then, if the right parallelepipeds AF, GO be constituted on the bases AC, GK, with the altitudes AE and GM, they will be equal to the parallelepipeds W, Z. Now by the first case the solid AF has to the solid GO the ratio compounded of the ratios of the base AC to the base GK, and of the altitude AE to the altitude GM; therefore also the oblique parallelepiped W has to the oblique parallelepiped Z the ratio compounded of the same ratios.

Wherefore parallelepipeds are to another, &c.: which was to be proved.

Scholium. Since arithmetically the ratio compounded of two ratios m to n and p to g is the ratio m x p to n x q (Alg. 293), if A and a be numbers proportional to the altitudes of two parallelepipeds P and p, and B and b numbers proportional to the areas of their bases, then by this proposition, P:p::AxB: axb.

Also since the parallelograms which are the bases of the solids are equal to rectangles having the same base and altitude as the parallelograms, and that these rectangles are to each other in the

ratio compounded of the ratios of their sides (VI. 23), if C and c be the bases, and D and d the altitudes of the parallelograms B and b, then B:b:: CxD : cxd P:p::AxCxD: axcxd.

and consequently

PROP. LIX. THEOR.

Similar parallelepipeds are to one another in the triplicate ratio of their homologous edges.

Let AG, KQ (fig. 67) be two similar parallelepipeds, of which AB and KL are two homologous sides: the ratio of the solid AG to the solid KQ is triplicate of the ratio of AB to KL.

Because the solids are similar, the parallelograms AF, KP are similar (Def. 13), as also the parallelograms AH, KR; therefore the ratios of AB to KL, AE to KO and AD to KN are all equal (VI. Def. 1). But the ratio of the solid AG to the solid KQ is compounded of the ratios of AC to KM and AE to KO; and the ratio of AC to KM is compounded of the ratios of AB to KL and AD to KN, because AC and KM are similar parallelograms (VI. 23); consequently the ratio of the solid AG to the solid KQ is compounded of the three ratios of AB to KL, AD to KN, and AE to KO. But these three ratios are equal; therefore the ratio that is compounded of them is triplicate of any one of them, and consequently the ratio of the parallelepiped AG to the parallelepiped KQ is the triplicate of the ratio of AB to KL. Wherefore similar parallelepipeds, &c. : which was to be proved.

Cor. As cubes are similar parallelepipeds, the cube on AB is to the cube on KL in the triplicate ratio of AB to KL, that is, in the same ratio as the solid AG to the solid KQ. Similar parallelepipeds are therefore to one another as the cubes on their homologous sides.

PROP. LX. THEOR.

If a prism be cut by parallel planes the sections are equal and similar plane figures.

Let the prism ABCDEFGHIK (fig. 68) be cut by the parallel planes LO and QT, the sections LMNOP and QRSTV are equal and similar plane figures.

Because the parallel planes LO and QT are cut by the plane AG, the sections LM and QR are parallel (Prop. 26); and LQ and MR are parallel (Def. 15): therefore LMRQ is a parallelogram, and LM is equal to QR. In like manner it may be shown that MN, NO, OP, PL are respectively parallel and equal to RS, ST, TV, VQ. And because LM and MN are parallel to QR and RS, the angle LMN is equal to the angle QRS (Prop. 20). In like manner it may be shown that the angles MNO, NOP, OPL, PLM are severally equal

to the angles RST, STV, TVQ, VQR; consequently the figures LMNOP and QRSTV are equiangular; and they have the sides about their equal angles equal each to each: they are therefore equal and similar. Wherefore if a prism be cut by parallel planes, &c. : which was to be proved.

PROP. LXI. THEOR.

If a prism, having a triangle for its base, and a parallelepiped, be upon equal bases and have equal altitudes, they shall be equal to one an

other.

Let the prism ACE (fig. 69) having the triangle ABC for its base, and the parallelepiped GH be upon equal bases ABC and GK, and have equal altitudes, namely the perpendiculars let fall upon the planes ABC and GK from any point in the opposite planes DEF and IH; the prism ACE is equal to the parallelepiped GH.

Bisect AC in L, draw BN parallel to AC, and LM, CN parallel to AB; complete the parallelepiped BF; and through LM let the plane LP pass parallel to AH, dividing the parallelepiped into two equal parallelepipeds BO, MF (Prop. 53. Cor.).

And because the prism ACE is half the parallelepiped BF (Prop. 54), and that the parallelepiped BO is also half the parallelepiped BF, the prism ACE is equal to the solid BO. But the parallelogram BL is equal to the triangle ABC (I. 42); and the parallelogram GK is by hypothesis equal to the triangle ABC; therefore the base GK is equal to the base BL: and the solids GH and BO have also equal altitudes; consequently they are equal (Prop. 56). But it has been shown that the prism ACE is equal to the solid BO; therefore the prism ACE is likewise equal to the parallelepiped GH. Wherefore if a prism having a triangle for its base, &c.: which was to be proved.

PROP. LXII. THEOR.

If a prism having any rectilineal figure for its base, and a parallelepiped be upon equal bases, and have the same altitude, they shall be equal to one another.

Let the prism ABCDEFGHIK (fig. 70) having the polygon ABCDE for its base, and the parallelepiped LM, be upon equal bases ABCDE and LN, and have equal altitudes; the prism ABCDEFGHIK is equal to the parallelepiped LM.

In the base of the prism join AC, AD, and in the opposite plane, FH, FI, dividing these planes into triangles. Describe the parallelogram OP equal to the triangle ABC (I. 42); to PQ apply the parallelogram QR equal to the triangle ACD and having the angle PQS equal to the angle UOQ (I. 44); and to RS apply the paral