Book XI. PROP. XXXIV. THEOR. THE bases and altitudes of equal solid parallelopi. See Note. peds, are reciprocally proportional; and if the bases and altitudes be reciprocally 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 NP be equal, but the alti Α E C N tudes AG, CM be not equal, 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 be not, neither also, in this case, R D K altitude CT. Because the solid AB is equal to the A E C N solid CD, therefore the solid AB is to the solid CV, as a the so- a 7. 5, 1 240 Book XI. lid CD to the solid CV. b is the base EH to the c 25. 11. But as the solid AB to the solid CV, so base NP; for the solids AB, CV are of b 32. 11. the same altitude; and as the solid CD to CV, so is the base MP to the base 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 parallelopipeds AB, CD are reciprocally proportional to their altitudes. d 1. 6. 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 alti- e P E C N tude of the solid CD is to the altitude of the solid AB, there e A. 5. fore the altitude of CD is equal to the altitude of AB. But solid parallelopipeds upon equal bases, and of the same altitude, are equal to one another: therefore the solid AB is equal to the solid CD. f 31.11 K 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 solid CD to AG the altitude of AB, CM is greater e than AG. Again, take CT equal to AG, and complete, as before, the solid CV. And because the base EH is to the H base NP, as CM to AG, R D B F G C N CT, therefore the base EH is to the base NP, as MC to CT. But as the base EH is to NP, so b 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 CV: and therefore as the Book XI. 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 © 25. 11. CV; and therefore the sold 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 solid BA, for they are upon the same base FK, and to the g 29. or of the 30. 11. same altitude; and that the solid DC is equals 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 Book XI. 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 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 before proved, the g 29. or solid BT is equal to the solid DZ: but BT is equals 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. 30. 11. Book XL PROP. XXXV. THEOR. IF, from the vertices of two equal plane angles, See Note. 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 anglè 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 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 a to the same plane: from the points a 8. 11. K, N, to the straight lines AB, AC, DE, DF, draw perpendiculars KB, KC, NE, NF; and join HB, BC, ME, EF: |