Book XI. 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. e 27. 11. Take AB to CD, as EF to ST, and from ST describe e a solid parallelepiped SV, similar and similarly situated to either of the solids EM, GN: And because A B is to CD, as EF to ST, and because from AB, CD the solid parallelepipeds AK, CL are similarly described; and in like manner, the solids EM, SV from the straight lines, EF, ST; therefore AK is to CL f 9. 5. as EM to SV: But by the hypothesis 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 are equal to one another: And because as AB to CD so is EF to ST, and ST is equal to GH; AB is to CD as EF to GH. Therefore, “if four straight “ lines," &c. Q. E. D. See N. 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; Book XI. “ and let it meet the plane AB in the point F; and from F “ drawa in the plane AB a perpendicular FG to DA, which is a 12. 1. “ also perpendicularb to the plane CD; and join EG: Then b 4. def.11. “ because FG is perpendicular “ to the plane CD, and the C E “ straight line EG which is in " that plane meets it; there“ fore FGE is a right angle: c 3. def. 11. “ But EF is also at right angles A D “ to the plane AB; and there“ fore EFG is a right angle: T 66 Wherefore two of the angles B o of the triangle EFG are equal “ together to two right angles; which is absurd: Therefore “ the perpendicular from the point E to the plane A B does “ not fall elsewhere than upon the straight line AD; it there“ fore falls upon it. If, therefore, a plane,'" &c. Q. E. D. PROP. XXXIX. THEOR. In a solid parallelepiped, 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 parallelepiped cut each other into two equal parts. Let the sides of the opposite planes D CF, AH of the solid 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 because DK, CL are equal and parallel, KL is parallela to B DC: For the same reason, MN is parallel to BA: And BA is parallel to Book XI. to DC; therefore because KL, BA are each of them parallel Sto DC, and not in the same plane with it, KL is parallel v to b 9.11. 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 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, c 29. 1. the alternate angles DXY, YOE are equal to one another: And because DX Y C d 4. 1. DY is equal d to E T OYE, and DYE is H P R A to SG: And be N G cause CA is equal and parallel to DB, and also equal and parallel to EG; theref 30. 1. fore DB is equal and parallel to EG: and DE, BG join a 33. 1. 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 manifest, that DG, YS must meet one another; let them meet in T: And because DE is parallel to BG, the alternate 8 15. ). angles EDT, BGT are equal o ; and the angle DTY, is equal & 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, h 26. 1. BG: Therefore the remaining sides are equal", each to each. Wherefore DT is equal to TG, and YT equal to TS. Wherefore, “ if in a solid,” &c. Q. E. D. Book XI. 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 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 the triangle GHK, the prism ABCDEF is equal to the prism GHKLMN. Complete the solids AX, GO; and because the parallelogram AF is double the triangle GHK; and the parallelogram B D M 0 HK double a the same triangle; therefore the parallelogram a 31. 1. AF is equal to HK. But solid parallelepipeds upon equal bases, and of the same altitude, are equal to one another. b 31. 11. Therefore the solid AX is equal to the solid GO; and the prism ABCDEF is halfċ the solid AX; and the prism c 28. 11. GHKLMN half the solid GO. Therefore the prism ABCDEF is equal to the prism GHKLMN. Wherefore, “ if there be two,” &c. Q. E. D. THE ELEMENTS OF EUCLID. . BOOK XII. LEMMA I. Which is the first proposition of the tenth book, and is necessary to some of the propositions of this book. Book XII. See N. If from the greater of two unequal magnitudes, So there be taken more than its half, and from the re mainder 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 D its balf, and so on; there shall at length remain A a magnitude less than C. For C may be multiplied so as at length to Kbecome greater than AB. Let it be so multiplied, and let DE its multiple be greater than -F AB, and let DE be divided into DF, FG, GE, Heach equal to C. From AB take BH greater than its half, and from the remainder AH -G 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 be AK, KH, HB; and the divisions in ED be DF, FG, GE: And because DE is greater than A B, and BCE |