therefore as the solid AK is to the solid CL, so is the solid FM to the solid HN (b). [2.] Next, let the solid AK be to the solid CL, as the solid FM is to the solid HN: the straight line AB shall be to CD, as EF is to GH. Take as AB is to CD, so is EF to ST, and from ST describe a solid parallelopiped SV similar and similarly situated to either of the solids FM, HN (e). And because AB is to CD, as EF is to ST, and that from AB, CD the solid parallelopipeds AK, CL are similarly described; and in like manner the solids FM, SV from the straight lines EF, ST; therefore AK is to CL, as FM is to SV; but, by the hypothesis, AK is to CL, as FM to HN; therefore HN is equal to SV (f): but it is likewise similar and similarly situated to SV; therefore the planes which contain the solids HN, SV are similar and equal, and their homologous sides GH, ST equal to one another: and because as AB is to CD, so is EF to ST, and that ST is equal to GH, therefore AB is to CD, as EF is to GH. PROPOSITION XXXVIII. THEOREM.-"If a plane (CD) be perpendicular to another plane (AB), and a straight line be drawn from a point (E) in one of the planes (CD) perpendicular to the other plane (AB), this straight line shall fall on the common section (AD) of the planes." DEMONSTRATION. E "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 (a), which is also perpendicular to the plane CD (b); 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 (c): but EF is also at right angles to the plane AB, and therefore EFG is a right angle: wherefore two of the angles of the triangle EFG are equal together to two right angles; which is absurd (d); therefore the perpendicular from the point E to the plane AB, does not fall elsewhere than upon the straight line AD; that is, it therefore falls upon it." (a) I. 12. PROPOSITION XXXIX. THEOREM.—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. B K M E DEMONSTRATION. Let the sides of the 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 to DC (a): for the same reason, 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 parallel to BA (b): and because KL, MN are each of them parallel to BA, and not in the same plane with it, KL is parallel to MN (6): 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 shall meet, and cut one another into two equal parts. (a) I. 33. (e) I. 14. (b) XI. 9. (f) I. 15. (c) I. 29. (g) I. 26. (d) I. 4. Join DY, YE, BS, SG. Because DX is parallel to OE, the alternate angles DXY, YOE are equal to one another (c): and because DX is equal to OE, and XY to YO, and that they contain equal angles, the base DY is equal to the base YE (d), and the other angles are equal; therefore the angle XYD is equal to the angle OYE, and DYE is a straight line (e): for the same reason, BSG is a 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 (6): and DE, BG join their extremities; therefore DE is equal and parallel to BG (a): 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 (c): and the angle DTY is equal to the angle GTS (f): 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 remaining sides are equal, each to each (g): wherefore DT is equal to TG, and YT equal to TS. PROPOSITION XL. THEOREM.-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. DEMONSTRATION. 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 LÍ, 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 shall be equal to the prism GHKLMN. Complete the solids AX, GO: and because the parallelogram AF is double of the triangle GHK, and the parallelogram HK double of the same triangle, therefore the parallelogram AF is equal to HK (a): but solid parallelopipeds upon equal bases, and of the same altitude, are equal to one another (b); there B E G (a) I. 34. (b) XI. 31. (c) XI. 28. fore the solid AX is equal to the solid GO: and the prism ABCDEF is half of the solid AX (c): and the prism GHKLMN half of the solid GO (c): therefore the prism ABCDEF is equal to the prism GHKLMN. THE ELEMENTS OF EUCLID. BOOK XII. LEMMA I. THEOREM.-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. A K F DEMONSTRATION. 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. For C may be multiplied so as at length to 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, 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 because DE is greater than AB, and that EG taken from DE is not greater than its half, but BH taken from AB is greater than its half, therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, but HK is greater than the half of HA; therefore the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK is less than C. COROLLARY. And if only the halves be taken away, the same thing may, in the same way, be demonstrated. SCHOLIUM. This is the first proposition in the 10th book, and being nocessary to some of the propositions of this book, it is here inserted. PROPOSITION I. THEOREM. Similar polygons (ABCDE, FGHKL) inscribed in circles, are to one another as the squares on their dia BAE, GFL having an V. Def. 10 and 22. VI. 20. angle in one, equal to an angle in the other, and the sides about the equal angles proportionals, are equiangular; and therefore the angle AEB is equal to the angle FLG: but AEB is equal to AMB, because they stand upon the same circumference (b): and the angle FLG is, for the same reason, equal to the angle FNG: therefore also the angle AMB is equal to FNG; and the right angle BAM is equal to the right angle GFN (c); wherefore the remaining angles in the triangles ABM, FGN are equal, and they are equiangular to one another; therefore as BM is to GN, so is BA to GF (d); and therefore the duplicate ratio of BM to GN, is the same with the duplicate ratio of BA to GF (e): but the ratio of the square on BM to the square on GN, is the duplicate ratio of that which BM has to GN; and the ratio of the polygon ABCDE to the polygon FGHKL is the duplicate of that which BA has to GF (ƒ): therefore as the square on BM is to the square on GN, so is the polygon ABCDE to the polygon FGHKL. |