VX 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 compounded (def. A. 5.) of the ratios of a to b, b to c, and c to d, which are the same with 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. PROP. XXXIV. THEOR. The bases and altitudes of equal solid parallelopipeds, are reciprocally proportional: and if the bases and altitudes be recipro. cally proportional, the solid parallelopipeds are equal.* Let AB, CD be equal solid parallelopipeds; 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 the solid AB is likewise equal to the solid CD, CM shall be equal to AG. Because if the bases EH, NP be LLP equal, but the altitudes AG, CM H be not equal, neither shall the A E CN solid AB be equal to the solid CD. But the solids are equal, by the hypothesis. Therefore the al. titude 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. * See Note. 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 if it R D be not, neither also, in this case, would the solids AB, K B CD be equal, which, by the hypothesis, are equal. Make then CT equal to AG, and complete the solid parallelopiped CV of which the base is NP, and the altitude CT. H olan Because the solid AB is equal to the solid CD, therefore the А CN solid AB is to the solid CV, as (7. 5.) the solid CD to the solid CV. But as 'the solid AB to the solid CV, so (32. 11.) is the base EH to the base NP; for the solids AB, CV are of the same altitude; and as the solid CD to CV, so (25. 11.) is the base MP to the base PT, and so (1. 6.) 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 parallelopipeds AB, CD are reciprocally proportional to their altitudes. Let now the bases of the solid parallelopipeds AB, CD be reciprocally proportional to their altitudes; viz. as the base EH to the base NP, so the altitude of the solid K B R D CD to the altitude of the solid AB; the solid AB is equal to the FM 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 H V NP, since EH is to NP, as the altitude of the solid CD is to the A E C 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 altitude of the R D solid CD to AG the altitude of AB, CM is greater (A. 5.) than AG. Again, к в 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 A E C N to the base NP, as MC to CT. But as the base EH is to NP, so (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 GD 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 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; Q, I, U, Z; and complete the solids FV, XU, which are parallelopipeds, 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 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 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, 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 that 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. But the alti. tudes 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 before proved, the solid BT is equal to the solid DZ: but BT is equal (29. or 30. 11.) to the solid BA, and DZ to the solid DC, because they are upon the same bases, and of the same altitude. Therefore the solid AB is equal to the solid CD. Q. E. D. PROP. XXXV. THEOR. IF, from the vertices of two equal plane angles, there be drawn two straight lines elevated above the planes in which the angles are, and containing equal angles with the sides of those angles, each to each; and if in the lines above the planes there be taken any points, and from them perpendiculars be drawn to the planes in which the first named angles are: And from the points in which they meet the planes, straight lines be drawn to the vertices of the angles first named; these straight lines shall contain equal angles with the straight lines which are above the planes of the angles.* Let BAC, EDF be two equal plane angles; and from the points A, D let the straight lines AG, DM be elevated above the planes of the angles, making equal angles with their sides, each to each, viz. the angle GAB equal to the angle MDE, and GAC to MDF: and in AG, DM let any points G, M be taken, and from them let * See Note. perpendiculars GL, MN be drawn to the planes BAC, EDF, meeting these planes in the points L, N, and join LA, ND: the angle GAL is equal to the angle MDN. Make AH equal to DM, and through H draw HK parallel to GL. But GL is perpendicular to the plane BAC; wherefore HK is perpendicular (8. 11.) to the same plane: from the points K, N to the straight lines AB, AC, DE, DF, draw perpendiculars KB, KC, NE, NF; and join HB, BC, ME, EF: Because HK is perpendicular to the plane BAC, the plane HBK which passes through HK is at right angles (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 perpendicular (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 E is equal to DF; therefore, since AB is equal to DE, BA and AC are equal to ED and DF; and the angle BAC is equal to the angle EDF; wherefore the base BC is equal (4. 1.) to the base EF, and the remaining angles to the remaining angles: the angle ABC is therefore |