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 DM, the square of AH is equal to the square of DM: but the squares of AK, KH are equal to the square (47. 1.) of AH, 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 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, each to each, and the base HK to the base MN, as has been proved; therefore the angle HAK is equal (8. 1.) to the angle 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. 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 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 MN (13. 11.), which is perpendicular to the plane EDF, because these planes coincide with one another. Therefore HK is equal to MN. Q. E. D. PROP. XXXVI. THEOR. If ihree 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 parallelopiped DH. Make LK equal to A, and at the point K in the straight line LK make (26. 11.) a solid angle contained by the three plane angles LKM, MKN, NKL, equal to the angles EDF, FDG, GDE, each to each; and make KN equal to B, and KM equal to C; and complete the solid parallelopiped K E А B C 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 reciprocally proportional; therefore the parallelogram LM is equal (14. 6.) 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 the points G, N, to the planes EDF, LKM are equal (Cor. 35. 11.) to one another: therefore the solids KO, DH are of the same altitude; and they are upon equal bases LM, EF, and therefore they are equal (31. 11.) to one another: but the solid 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 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. * * See Note. 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 described from them. AK is to CL, as EM to GN. Make (11. 6.) AB, CD, O, P continual proportionals, as also EF, GH, Q, R; and because as, AB is to CD, so EF to GH; and that F G H Q R CD is (11. 5.) to O, as GH to Q, and O to P, as Q to R; therefore, ex æquali (22. 5.), AB is to P, as EF to R: but as AB to P, so (Cor. 33. 11.) is the solid AK to the solid CL; and as EF to R, so (Cor. 33. 11.) is the solid EM to the solid GN; therefore (11. 5.) as the solid AK to the solid CL, so is the solid EM to the solid GN. But let the solid AK be to the solid CL, as the solid EM to the solid GN: the straight line AB is to ED, as EF to GH. Take AB to CD, as EF to ST, and from ST describe (27. 11.) a solid parallelopiped SV similar and similarly situated to either of the solids EM, GN: and because AB is to CD, as EF to T F G H Q R ST, and that from AB, CD the solid parallelopipeds AK, CL are similarly described, and in like manner the solids EM, SV from the straight lines EF, ST; therefore AK is to CL, as EM to SV: but, by the hypothesis, AK is to CL, as EM to GN; therefore GN is equal (9. 5.) 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. “ 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 their 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. “For, if it does not, let it, if possible, fall elsewhere, as EF; and let it meet the plane AB in the point F; and from F draw (12. 1.) in the plane AB a perpendicular FG to DA, which is also perpendicular (4. def. 11.) to the plane CD; and join EG: then because FG is perpendicular to the plane CD, and the C straight line EG, which is in that plane, meets it; therefore FGE is a right angle (3. def. 11.): but EF is also at right angles to the plane AB; and therefore a EFG is a right angle: wherefore two of the angles of the triangle EFG are equal together to two right angles; B 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. PROP. XXXIX. THEOR. In a solid parallelopiped, if the sides of two of the 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.* * See Note. Y No Let the sides of the D K opposite planes CF, AH of the solid parallelopiped 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, CL are equal and parallel, KL is parallel (33. 1.) to DC: for the same rea- B son, MN is parallel to BA: and BA is parallel to DC; therefore because KL, BA, are Α Ν each of them parallel to DC, and not in the same plane with it, KL is parallel (9. 11.) to BA : and because KL, MN are each of them parallel to BA, and not in the same plane with it, KL is parallel (9. 11.) to MN; wherefore KL, MN are in one plane. In like manner, it may be proved, that XO, PR are in one plane. Let YS be the common section of the planes KN, XR; and DG the diameter of the solid parallelopiped AF: YS and DG do meet, and cut one another into two equal parts. Join DY, YE, BS, SG. Because DX is parallel to OE, the alternate angles DXY, YOE are equal (29. 1.) to one another: and because DX is equal D K to OE, and XY to YO, and contain equal an Y gles, the base DY is equal (4. 1.) to the base YE, and the other angles are equal; therefore the angle XYD is equal to the angle OYE, and DYE is a straight (14. 1.) line: for the same reason BSG is a straight line, and BS equal to SG: and because CA SV is equal and parallel to DB, and also equal A and parallel to EG, therefore DB is equal and parallel (9. 11.) to EG: and DE, BG join their extremities; therefore DE is equal and parallel (33. 1.) to BG: and DG, YS are drawn from points in the one, to points in the other: and are therefore in one plane: whence it is manifest, that DG, YS must meet one another; let them meet in T: and because DE is parallel to BG, the alternate angles EDT, BGT are equal (29. 1.); and the angle DTY is equal (15. 1.) to the angle GTS: therefore in |