BOOK XI. • 27. 11. But let the solid AK be to the solid CL, as the solid EM to the solid GN: The straight line AB is to CD, as EF to GH. Take AB to CD, as EF to ST, and from ST describe 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 the solids EM, SV from the straight lines EF, ST: there 鸭鸭 E G H Q R 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 equal 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 66 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 AĎ. 66 66 A E Book XI. 12. 1. 4 Def. 11. ID $3 Def. 11. "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, in the plane AB, a perpendicular FG to DA, which is also perpendicular to the plane CD; 4 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 angle: But "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. PROP. XXXIX. THEOR. B 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 twoequal parts. BOOK XI. son, MN is parallel to BA: And BA is parallel to DC: therefore, because KL, BA are each of them parallel to DC, and not in the same plane with it, KL is paralb 9. 11. lel to BA: And because KL, MN are each of them parallel to BA, and not in the same plane with it, KL is parallel 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, © 29. 1. the alternate angles DXY, YOE are equal to one another: : For the same rea- straight line, and BS equal to SG: And because CA is equal and parallel to DB, and also equal and parallel to EG; therefore DB is equal and parallel to EG: And DE, BG join their extremities; therefore DE is equal and 33. 1. parallela 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 pa rallel to BG, the alternate angles EDT, BGT are equal: 15. 1. and the angle DTY is equalf to the angle GTS: Therefore in the triangles DTY, GTS there are two angles in the one equal to two angles in the other, and one side equal to one side, opposite to two of the equal angles, viz. DY to GS; for they are the halves of DE, BG: Therefore the re26. 1. maining sides are equals, each to each. Wherefore DT is equal to TG, and YT equal to TS. Wherefore, if in a solid, &c. Q. E. D. · PROP. XL. THEOR. If there be two triangular prisms of the same altitude, the base of one of which is a parallelogram, and the base of the other a triangle; if the parallelogram be double of the triangle, the prisms shall be equal to one another. Let the prisms ABCDEF, GHKLMN be of the same altitude, the first 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 base; if the parallelogram AF be double of the triangle GHK, the prism ABCDEF is equal to the prism GHKLMN. Complete the solids AX, GO; and because the parallelogram AF is double of the triangle GHK; and the paral Book XI. 31. 11. lelogram HK doublea of the same triangle; therefore the 34. 1. parallelogram AF is equal to HK. But solid parallelopipeds upon equal bases, and of the same altitude, are equalb to one another. Therefore the solid AX is equal to the solid GO; and the prism ABCDEF is half of the solid AX; and the prism GHKLMN half of the solid GO. Therefore the prism ABCDEF is equal to the prism GHKLMN. Wherefore, if there be two, &c. Q. Ë. D. 28. 11. 244 THE ELEMENTS OF EUCLI D. BOOK XII. LEMMA I. BOOK XII. Which is the first proposition of the tenth book, and is necessary to some of the propositions of this book. See N. IF from the greater of two unequal magnitudes, there be taken more than its half, and from the remainder more than its half; and so on: There shall at length remain a magnitude less than the least of the proposed 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 so on; there shall at length remain a magnitude less than C. A For C may be multiplied so as at length to K become greater than AB. Let it be so multiplied, and let DE its multiple be greater, than AB, and let DE be divided into DF, FG, H GE, each equal to C. From AB take BH greater than its half, and from the remainder AH take HK greater than its half, and so on, until there be as many divisions in AB as there are in DE: And let the divisions in AB F G D be AK, KH, HB; and the divisions in ED B CE be DF, FG, GE. And because DE is greater than AB, and |