Book XI. € 27. 11. But let the folid AK be to the folid CL, as the folid EM to the folid GN: The ftraight line AB is to CD, as EF to GH. Take AB to CD, as EF to ST, and from ST defcribe ea folid parallelepiped SV fimilar and fimilarly fituated to either of the folids EM, GN: And becaufe AB is to CD, as EF to ST, and that from AB, CD the folid parallelepipeds AK, CL are fimilarly defcribed; and in like manner the folids EM, SV from the ftraight lines EF, ST; therefore AK is to CL, aş £9. 5. See N. E F G H Q R KM to SV: But, by the hypothefis, AK is to CL, as EM to GN: Therefore GN is equal fto SV: But it is likewife fimilar and fimilarly fituated to SV; therefore the planes which contain the folids GN, SV are fimilar and equal, and their homologous fides GH, ST equal to one another: And because as AB to CD, fo EF to ST, and that ST is equal to GH; AB is to CD, as EF to GH. Therefore, if four traight lines, &c. Q. E. D, ' PRO P. XXXVIII. THEOR. IF a plane be perpendicular to another plane, and a ftraight line be drawn from a point in one of "the planes perpendicular to the other plane, this ftraight line fhall fall on the common fection of the "planes. "Let the plane CD be perpendicular to the plane AB, and "let AD be their common fection; if any point E be taken in "the plane CD, the perpendicular drawn from E to the plane AB fhall fall on AD. * For C "For, if it does not, let it, if poffible, fall elsewhere, as EF; Book XI. " and let it meet the plane AB in the point F; and from F "draw 2, in the plane AB, a perpendicular FG to DA, which a 12, 1. " is also perpendicular b to the plane CD; and join EG: Then b 4. def.z z. "becaufe FG is perpendicular "to the plane CD, and the "ftraight line EG, which is in "that plane, meets it; there"fore FGE is a right angle : "But EF is alfo at right angles "to the plane AB; and there"fore EFG is a right angle: "Wherefore two of the angles A E G 3. def. 13. D B "of the triangle EFG are equal together to two right angles; 66 PRO P. XXXIX. THEOR. IN a folid parallelepiped, if the fides of two of the Sce Ne Let the fides of the oppofite planes CF, AH of the folid parallelepiped 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 becaufe DK, CL are equal and parallel, KL is parallel a to DC: For the fame reafon, MN is parallel to BA: And b. 9. 11. Book XI. BA is parallel to DC; therefore becaufe KL, BA are each of them parallel to DC, and not in the fame plane with it, KL is parallel b to BA: And because KL, MN are each of them parallel to BA, and not in the fame plane with it, KL is parallel b 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 fection of the planes KN, XR; and DG the diameter of the folid parallelepiped 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 to one another : And because DX is e- D qual to OE, and K F SG: And because CA is equal and parallel to DB, and alfo e. qual and parallel to EG; therefore DB is equal and parallel û to EG: And DE, BG join their extremities; therefore DE is equal and parallel a 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 manifeft, that DG, YS muft meet one another; let them meet in T: And becaufe DE is parallel to BG, the alternate angles EDT, BGT are equal; and the angle DTY is equal f to the angle GTS: Therefore in the triangles DIY, GTS there are two angles in the one equal to two angles in the other, and one fide equal to one fide, oppofite to two of the equal angles, viz. DY to GS; for they are the halves of DE, BG: Therefore the remaining fides are equal g, each to each. Wherefore DT is equal to TG, and YT equal to TS. Therefore, if in a folid, &c. Q. E, D. PROP. XL. THEOR. IF there be two triangular prifms of the fame alti tude, the bafe of one of which is a parallelogram, and the bafe of the other a triangle; if the parallelogram be double of the triangle, the prifius fhall be equal to one another. Let the prifms ABCDEF, GHKLMN be of the fame altitude, the firft whereof is contained by the two triangles ABE, CDF, and the three parallelograms AD, DE, EC; and the other by the two triangles GHK, LMN and the three parallelograms LH, HN, NG; and let one of them have a parallelogram AF, and the other a triangle GHK for its bafe; if the parallelogram AF be double of the triangle GHK, the prifm ABCDEF is equal to the prifm GHKLMN. Complete the folids AX, GO; and because the parallelogram AF is double of the triangle GHK; and the parallelo Book XI. gram HK double a of the fame triangle; therefore the paral-a 34. 1. lelogram AF is equal to HK. But folid parallelepipeds upon Dequal bafes, and of the fame altitude, are equal b to one an-b 31. 11. other. Therefore the folid AX is equal to the folid GO; and the prifm ABCDEF is half of the folid AX; and the prifme 18. 11. GHKLMN half of the folid GO. Therefore the prifm ABCDEF is equal to the prifm GHKLMN. Wherefore, if there be two, &c. Q. E. D. THE Which is the first propofition of the tenth book, and is neceffary to fome of the propofitions of this book. IF from the greater of two unequal magnitudes, there be taken more than its half, and from the remainder more than its half; and fo on: There fhall at length remain a magnitude lefs than the leaft of the propofed magnitudes. Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken more than its half, and from the remainder more than its half, and fo on; there fhall at length remain a magnitude lefs than C. D For C may be multiplied fo, as at length to become greater than AB. Let it be fo multiplied, and let DE its multiple be greater than F AB, and let DE be divided into DF, FG, GE, H G take HK greater than its half, and fo on, until KH, HB; and the divifions in DE be DF, FG, B CE that |