Book XL. 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. 1 Let AB, CD be equal solid parallelopipeds; their bases are First, Let the insisting straight lines, AG, EF, LB, HK; G M X Р H A E CAN Next, Let the bases EH, NP not be equal, but EH greater R D K M F T L P А E C N and altitude CT. Be... cause the solid AB is equal to the solid CD, therefore the solid AB is to the solid CV, as the solid CD to the solid a 7.5. N Book XI. CV. But as the solid AB to the solid CV, sob is the base EH to the base NP; for the solids AB,'CV are of the same b 32 Il. altitude; and as the solid CD to CV, soc is the base MP to c 25. 11. the base PT, and so d is the straight line MC to CT; and L. 6. CT is equal tu AG. Therefore, as the base EH to the base NP, so is MC to AG. Wherefore, the bases of the solid pa. rallelopipeds 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 CD to K B R D the altitude of the solid G F M X P o A E с the altitude of the solid CD is to the altitude of the solid AB, e A. S. therefore the altitude of CD is equal e to the altitude of AB. But solid parallelopipeds upon equal bases, and of the same 31. 11. altitude, are equal i 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 R D B M X than AG. Again, take K F G T L 0 base NP, as CM to AG, H Р C N e V VT, 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; in that is, each of the solids AB, CD has the same rațio to the c 25. 11. 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 pa-, rallelopipeds, 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 equals to the g 29. or 30. solid BA, for they are upon the same base FK, and of the 11. K B R D 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 pro: portional 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 solid BT is equal to the 8 29. or 30. solid DZ:but BT is equals to the solid BA, and DZ to the solid 11, DC, because they are upon the same bases, and of the same alti. tude. Therefore the solid AB is equal to the solid CD.Q. E. D. Book XI, PROP. XXXV. THEOR. . . IF, from the vertices of two equal plane angles, See Noto. 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 Ą, D let the straight lines AG, DM be elevated above the plants 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 ta. ken, 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. D G Make AH equal to DM, and through H draw IIK parallel to GL: but GL is perpendicular to the plane BAC; wherefore HK is perpendicular a to the same plane : from the points à 8. 11. K, N, to the straight lines AB, AC, DE, DF. draw perpendiculars KB, KC, NE, NF; and join HB, BC, ME, EF: |