Book XI. of which the base is NP, and altitude CT. Because the solid ad AB is equal to the solid CD, therefore the solid AB is to the soa 7. 5. lid CV, as a the solid CD to the solid CV. But as the solid AB b 32. 11. to the solid CV, so b is the base EH to the base NP; for the so lids AB, CV are of the same altitude; and as the solid CD to c 25. 11. CV, so c is the base MP to the base PT, and so d is the straight d 1. 6. 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 parallelepipeds AB, CD are reciprocally proportional to their altitudes. Let now the bases of the solid parallelepipeds 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 the K B R D altitude of the solid AB; G F M the solid AB is equal to X P A 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 sos lid parallelepipeds upon equal bases, and of the same altitude, f 31. 11. are equalf 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 alti RD х CM is greater e than B M F T L Р A E CN 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 c 25. 11. 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 parallelepipeds, 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 8 to the g 29. or solid BA, for they are upon the same base FK, and of the 30. 11. same altitude; and that the solid DC is equal 8 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 parallelepipeds 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 parallelepipeds 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 solid BT is equal to the solid DZ: but BT is equal 8 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. 8 29. or 30. 11. Book XI. PROP. XXXV. THEOR. IF from the vertices of two equal plane angles there See N. 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 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. u K, N, to the straight lines AB, AC, DE, DF, draw perpendiculars KB, KC, NE, NF; and join HB, BC, ME, EF: Book XI. because HK is perpendicular to the plane BAC, the plane HBK which passes through HK is at right angles b to the plane b 18. 11. BAC; and AB is drawn in the plane BAC at right angles to the common section BK of the two planes; therefore AB is c 4. def. perpendicular c to the plane HBK, and makes right angles d 11. with every straight line meeting it in that plane: but BH meets d 3. def. 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 $26. 1. are equale, each to each: wherefore AB is equal to DĒ. 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; and the angle BAC is equal to the angle EDF; wherefore the base BC is equalf to the base EF, and the remaining angles to the remaining angles: the angle ABC is therefore equal 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 angle 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 |