Kitap 11
37
Let AB, CD, EF, GH be four straight lines in proportion, so that, as AB is to CD, so is EF to GH; and let there be described on AB, CD, EF, GH the similar and similarly situated parallelepipedal solids KA, LC, ME, NG; I say that, as KA is to LC, so is ME to NG.
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For, since the parallelepipedal solid KA is similar to LC, therefore KA has to LC the ratio triplicate of that which AB has to CD. [XI. 33]
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For the same reason ME also has to NG the ratio triplicate of that which EF has to GH. [id.]
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And, as AB is to CD, so is EF to GH.
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Therefore also, as AK is to LC, so is ME to NG.
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Next, as the solid AK is to the solid LC, so let the solid ME be to the solid NG; I say that, as the straight line AB is to CD, so is EF to GH.
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For since, again, KA has to LC the ratio triplicate of that which AB has to CD, [XI. 33] and ME also has to NG the ratio triplicate of that which EF has to GH, [id.] and, as KA is to LC, so is ME to NG, therefore also, as AB is to CD, so is EF to GH.
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Therefore etc. Q. E. D.
PROPOSITION 38.
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If the sides of the opposite planes of a cube be bisected, and planes be carried through the points of section, the common section of the planes and the diameter of the cube bisect one another.
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For let the sides of the opposite planes CF, AH of the cube AF be bisected at the points K, L, M, N, O, Q, P, R, and through the points of section let the planes KN, OR be carried; let US be the common section of the planes, and DG the diameter of the cube AF.
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I say that UT is equal to TS, and DT to TG.
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For let DU, UE, BS, SG be joined.
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Then, since DO is parallel to PE, the alternate angles DOU, UPE are equal to one another. [I. 29]
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And, since DO is equal to PE, and OU to UP, and they contain equal angles, therefore the base DU is equal to the base UE, the triangle DOU is equal to the triangle PUE, and the remaining angles are equal to the remaining angles; [I. 4] therefore the angle OUD is equal to the angle PUE.
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For this reason DUE is a straight line. [I. 14]
