solid AB to the solid EX, so is EX to PL, and PL to KO: But if four magnitudes be continual proportionals, the first is said to have to the fourth the triplicate ratio of that which it has to the second: Therefore the solid AB has to the solid KO, the triplicate ratio of that which AB has to EX: But as AB is to EX, so is the parallelogram AG to the parallelogram GK, and the straight line AE to the straight line EK. Wherefore the solid AB has to the solid KO, the triplicate ratio of that which AE has to EK. And the solid KO is equal to

the solid CD, and the straight line EK is equal to the straight line CF. Therefore the solid AB has to the solid CD, the triplicate ratio of that which the side ÀE has to the homologous side CF, &c. Q. E. D.

COR. From this it is manifest, that, if four straight lines be continual proportionals, as the first is to the fourth, so is the solid parallelopiped described from the first to the similar solid similarly described from the second; because the first straight line has to the fourth the triplicate ratio of that which it has to the second.

Let AB, CD be solid parallelopipeds, of which AB is contained by the parallelograms AE, AF, AG equiangular, each to each, to the parallelograms CH, CK, CL, which contain the solid CD. The ratio which the solid AB has to the solid CD, is the same with that which is compounded of the ratios of the sides AM to DL, AN to DK, and AO to DH.

Produce MA, NA, OA to P, Q, R, so that AP be equal to DL, AQ to DK, and AR to DH; and complete the solid parallelopiped AX contained by the parallelograms AS, AT, AV similar and equal to CH, CK, CL, each to each. Therefore the solid AX is equal (C. 11.) 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 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 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.

titude of the solid AB; the solid AB is equal to the solid CD. Let the insisting lines be, as before, at right angles to the bases. Then, if the base EH be equal to the base NP, since EH is to NP, as the altitude of the Bolid CD is to the altitude of the solid AB, therefore the altitude of CD is equal (A. 5.) to the altitude of AB. But solid parallelopipeds upon equal bases, and of the same altitude, are equal (31. 11.) 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

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; Q, I, U, Z; and complete the solids FV, XU, which are parallelopipeds, as was proved in the last part of Prop. XXXI. 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 that the solid BT is equal (29. or 30. 11.) to the solid BA, for they are upon the same base FK, and of the

titude of the solid CD to AG the altitude of AB, CM is greater (A. 5.) than AG. Again, take CT equal to AG, and complete, as before, the solid CV. And, because the base EH is to the base NP, as CM to AG, and that AG is equal to CT, therefore the base EH is to the base NP, as MC to CT. But as the base EH is to NP, 20 (32. 11.) 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 MP to the base PT, and the solid CD to the solid (25. 11.) CV; And therefore as the solid AB to the solid CV, so is the 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 FF, BL, GA, KH; XN, DO, MC, RP not be at right angles to the bases of the solide; and from the points F, B, K, G; X,

same altitude; and that the solid DC is equal (29. or 30. 11.) 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 parallelopipeds 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 of the solid CD to the altitude of the solid AB; that is, the bases of the solid parallelopipeds AB,

(18. 11.) to the plane BAC; and AB is drawn in the plane BAC at right angles to the common section BK of the two planes: Therefore AB is per pendicular (4. def. 11.) to the plane HBK, and makes right angles (3. def. 11.) with every straight line meeting it in that plane: But BH meets it in that plane; therefore ABH is a right angle: For the same reason, DEM is a right angle, and is therefore equal to the angle ABH: And the angle HAB is equal to the angle MDE. Therefore in the two triangles HAB, MDE there are two angles in one equal to two angles in the other, each to each, and one side equal to one side, opposite to one of the equal angles in each, viz. HA equal to DM; therefore the remaining sides are equal, (26. 1.) each to each: Wherefore AB is equal to DE. In the same manner, if HC and MF be joined, it may be demonstrated that AC is equal to DF: Therefore, since AB is equal to DE, BA and AC are equal to ED and DF;

B

A

and the angle BAC is equal to the angle FDF; wherefore the base BC is equal (4. 1.) to the base EF, and the remaining angles to the remaining an gles: The angle ABC is therefore e qual to the angle DEF: And the right angle ABK is equal to the right angle DEN, whence the remaining angle CBK is equal to the remaining angle FEN: For the same reason, the an

gle BCK is equal to the angle EFN: Therefore in the two triangles BCK, EFN, there are two angles in one equal to two angles in the other, each to each, and one side equal to one side adjacent to the equal angles in each, viz. BC equal to EF; the other sides, therefore, are equal to the other sides; BK then is equal to EN: And AB is equal to DE; wherefore AB, BK are equal to DE, EN; and they contain right angles; wherefore the base AK is equal to the base DN: And since AH is equal to DM, the square of AH is equal to the square of DM: But the squares of AK, KH are equal to the square (47. 1.) of AH, because AKH is a right angle: And the squares of DN, NM are equal to the square of DM, for DNM is a right angle: Wherefore the squares of AK, KII are equal to the squares of DN, NM; and of those the square of AK is equal to the square of DN: Therefore the remaining square of KH is equal to the remaining square of NM; and the straight line KH to the straight line NM: And because HA, AK are equal to MD, DN, each to each, and the base HK to the base MN, as has been proved; therefore the angle HAK is equal (8. 1.) to the angle MDN. Q. E.D.

COR. From this it is manifest, that if, from the vertices of two equal plane angles, there be elevated two equal straight lines containing equal angles with the sides of the angles, each to each; the perpendiculars drawn from the extremities of the equal straight lines to the planes of the first angles are equal to one another.

qual to MDE, and HAC equal to the angle MDF; and from H, M let HK, MN be perpendiculars to the planes BAC, EDF: HK is equal to MN.

Because the solid angle at A is contained by the three plane angles BAC