PD X K B G A CN * 32. 11. but the altitudes AG, CM be not equal, neither shall the solid AB be equal to the solid CD: but the solids are equal, by the hypothesis; therefore the altitude 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. Next, let the bases EH, NP not be equal, but EH greater than the other: then since 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 would the solids AB, CD be equal, which, by the hypothesis, are equal. Make P then CT equal to AG, and complete the solid parallelopiped CV, of which the base is NP, and altitude CT. Because the solid AB is equal to the solid CD, therefore the solid AB is to the solid CV, as* the solid CD to the *7.5. solid CV: but as the solid AB to the solid CV, so 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* is the base MP to the base PT, and so is the * 25. 11. straight line MC* to CT; and CT is equal to AG: *1.6. 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 is to the base NP, so let GM be to AG: the solid AB shall be equal to the solid CD. If the base EH be equal to the base NP, then since EH is to NP, K B R D as the altitude of the solid CD is to the altitude of the solid AB, P therefore the altitude of CD is equal* Е 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 * 51. 11. 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 EĚ to the base NP, so is CM the altitude of the solid CD to AG the altitude of AB, CM is greater* * A. 5. than AG. Therefore, as before, take CT equal to AG, and complete the solid CV. And because the base EH FNM A * A. 5. * 32. 11. A CN † 1. 6. * 25, 11. 1 is to the base NP, as CM to AG, R D К M P! of the solids AB, CD has the same ratio to the solid + 9.5. CV; and therefore the solid AB is equal t to the solid Second general case. Let the insisting straight lines In this case, likewise, if the solids AB, CD be equal, their bases shall be reciprocally proportional to their altitudes, viz. the base ÉH to the base NP, as the altitude of the solid CD to the altitude of the solid AB. 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. Because the solid AB is equal to the solid CD, and * 29.or 30. that the solid AB is equal * to the solid BT, for they the same base F G * 29. or 30. CD is equal * to the solid DZ, being upon the same HIE s P 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 EHto the base NP, so is the altitude 11. are upon B R D 11. 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 shall be 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 K K B R D altitudes of the solids AB, F M BT are the same, as also of CD and DZ; therefore IY as the base FK to the base H SP 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 * to the solid BA, 29. or 30. and DZ to the solid DC, because they are upon the 11. same bases, and of the same altitude; therefore the solid AB-is equal to the solid CD. . Therefore, the bases, &c. Q. E. D. PROP. XXXV. THEOR. If, from the vertices of two equal plane angles, there be See N. 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 F BA K ES H 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 drawnt to the planes BAC, M plane BAC at right angles to the common section BK * 4 Def. 11. of the two planes; therefore AB is perpendicular* to *3 Def. 11, the plane HBK, and makes right angles * 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 + Hyp. equal to the angle ABH: and the angle HAB is equalt 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*, 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 + Hyp. and DF each to each; and the angle BAC is equal to the angle EDF: wherefore the base BC is equal * to the base EF, and the remaining angles to the remaining angles: therefore the angle ABC is equal to 18. 11. 26.1. 4. 1. 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 ECAKE 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; therefore the other sides 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, each to each; 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* of AH, * 47.1. 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, KH are equal to the squares of DN, NM: and of these 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* to the angle MDN. Therefore, if from the *8. 1. vertices, &c. ' 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. Another Demonstration of the Corollary. Let the plane angles BAC, EDF be equal to one another, and let AH, DM, be two equal straight lines above the planes of the angles, containing equal angles with BA, AC, ED, DF, each to each, viz. the angle |