AH is equal to DM, the square of AH is equal to the square Book XI. of DM: But the squares of AK, KH are equal to the square of AH, because AKH is a right angle: And the 8 47. 1. squares of DN, NM are equal to the square of DM, for DNM is a right angle : Wherefore the squares of AR, KH are equal to the squares of DN, NH; and of those 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, cach to each, and the base HK to the base MN, as has been proved: therefore the angle FAK is equalh to the angle h 8. 1. MDN. 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. dnother Demonstration of ihe 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 HAB equal to MDE, and HAC equal to the angle MDF; and from H, M, let HK, MN be perpendiculars to the planes BAC, EDF; HK is equal to MN. Because the solid angle at A is contained by the three plane angles BAC, BAH, HAC, which are, each to each, equal to the three plane angles EDF, EDM, MDF, containing the solid angle at D; the solid angles at A and D ,. are equal, and therefore coincide with one another; to wit, if the plane angle BAC be applied to the plane angle EDF, the straight line AH coincides with DM, as was shown in Prop. B. of this Book: And because AH is equal to DM, the point H coincides with the point M: Wherefore HK, which is perpendicular to the plane BAC, coincides with MNi which is perpendicular to the plane EDF, because i 13. 11. these planes coincide with one another. Therefore HK is equal to MN. Q. E. D. : Book XI. PROP. XXXVI. THEOR. See N. If three straight lines be proportionals, the solid parallelopiped described from all three as its sides, is equal to the equilateral parallelopiped described from the mean proportional, one of the solid angles of which is contained by three plane angles equal, each to each, to the three plane angles containing one of the solid angles of the other figure. Let A, B, C be three proportionals, viz. A to B, as B to C: The solid described from A, B, C is equal to the equilateral solid described from B, equiangular to the other. Take a solid angle D contained by three plane angles EDF, FDG, GDE; and make each of the straight lines ED, DF, DG equal to B, and complete the solid parallel opiped DH: Make LK equal to A, and at the point K in * 26. 11. the straight line LK, makea a solid angle contained by the three plane angles LKM, MKN, NKL, equal to the angles B T A EDF, FDG, GDE, each to each; and make KN equal to B, and KM equal to C; and complete the solid parallelopiped KO; And because, as A is to B, so is B to C, and that A is equal to LK, and B to each of the straight lines DE, DF, and C to KM; therefore LK is to ED, as DF to KM; that is, the sides about the equal angles are recipro1 14. 6. cally proportional; therefore the parallelogram LMis equal to EF; and because EDF, LKM are two equal plane angles, and the two equal straight lines DG, KN are drawn from their vertices above their planes; and contain equal angles with their sides, therefore the perpendiculars from Cor. 35, the points G, N, to the planes EDF, LKM are equals to 11. one another: Therefore the solids KO, DH are of the same Boox XI. altitude: and they are upon equal bases LM, EF, ard therefore they are equal to one another: But the solid " $1. 11. KO is described from the three straight lines A, B, C, and the solid DH from the straight line B. If therefore three straight lines, &c. Q. E. D. PROP. XXXVII. THEOR. If four straight lines be proportionals, the similar See N. solid parallelopipeds similarly described from them shall also be proportionals. And if the similar parallelopipeds similarly described from four straight lines be proportionals, the straight lines shall be proportionals.. Let the four straight lines AB, CD, EF, GH be proportionals, viz. as AB to CD, so EF, to GH; and let the similar parallelopipeds AK, CL, EM, GN be similarly des scribed from them. AK is to CL, as EM to GN. Makea AB, CD, O, P continual proportionals, as also. 11. 6. EF, GH, Q, R: And because as AB is to CD, so EF to P GH: and that CD is b to O, as GH to Q, and O to P, as Q 011.5. to R; therefore, ex æqualis, AB is to P, as EF to R: But c 22. 5. as AB to P, sort is the solid AK, to the solid CL; and as Cor. 33. EF to R, sod is the solid EM to the solid GN: Thereforeb 11. as the solid AK to the solid CL, so is the solid EM to the solid GN. Book XI. But let the solid AK he to the solid CL, as the solid m EM to the solid GN: The straight line AB is to CD, as EF to GH. • 27. 11. Take AB to CD, as EF to ST, and from ST describe e a solid parallelopiped SV similar and similarly situated to either of the solids EM, GN: And because AB is to CD, as EF to ST, and that from AB, CD the solid parallelopipeds AK, CL are similarly described ; and in like manner ihe solids EM, SV from the straight lines EF, ST: there E 7. B G T Ř fore AK is to CL, as EM to SV: But by the hypothesis, 19. 5. AK is to CL, as EM to GN: Therefore GN is equalf to SV: But it is likewise similar and similarly situated to SV; therefore the planes which contain the solids GN, SV are similar and equal, and their homologous sides GH, ST equal to one another : And because as AB to CD, so EF to ST, and that ST is equal to GH: AB is to CD, as EF to GH. Therefore, if four straight lines, &c. Q. E. D. PROP. XXXVIII. THEOR. See N.“ If a plane be perpendicular to another plane, and “a straight line be drawn from a point in one of “the planes perpendicular to the other plane, this “ straight line shall fall on the common section of “the planes.” “Let the plane CD be perpendicular to the plane AB, " and let AD be the common section; if any point E be “ taken in the plane CD, the perpendicular drawn from E « to the plane AB shall fall on AD. - 6 För, if it does not, let it, if possible, fall elsewhere, as Book' XI. “ EF; and let it meet the plane AB in the point F, and are " from F drawa, in the plane AB, a perpendicular FG to * 12. 1. DA, which is also perpendicular b to the plane CD; 04 Def. 11. fb and join EG: Then because «FG is perpendicular to the “plane CD, and the straight “line EG, which is in that plane, meets it; therefore “ FGE is a right anglec: But og Def. 11. “ EF is also at right angles to the plane AB; and there“ fore EFG is a right angle: “ Wherefore two of the angles of the triangle EFG are “equal together to two right angles; which is absurd: « Therefore the perpendicular from the point E to the “plane AB, does not fall elsewhere than upon the straight “ line AD; it therefore falls upon it. If therefore a plane,” &c. Q.E. D. B PROP. XXXIX. THEOR..! In a solid parallelopiped, if the sides of two of the See N. opposite planes be divided, each into two equal parts, the common section of the planes passing through the points of division, and the diameter of the solid parallelopiped, cut each other into two. equal parts. Let the sides of the opposite planes CF, AH, of the solid parallel opiped AF, be divided each into two equal parts in the points K, L, M, N; X, O, P, R; and join KL MN, XO, PR: And because DK, BL CL are equal and parallel, KL is parallel a to * DC: cont * 33. I. For the same rea |