第11卷命题 38 · 立方体截面与直径互相平分

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. I say that UT is equal to TS, and DT to TG. For let DU, UE, BS, SG be joined.

连接DU、UE、BS、SG。由于DO平行于PE，内错角相等，且DO等于PE、OU等于UP，夹角相等，故三角形DOU全等于三角形PUE，得DU等于UE，角OUD等于角PUE，因此DUE为直线。

Then, since DO is parallel to PE, the alternate angles DOU, UPE are equal to one another. [I. 29] 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. For this reason DUE is a straight line.

同理，BSG为直线，且BS等于SG。

[I. 14] For the same reason, BSG is also a straight line, and BS is equal to SG. Now, since CA is equal and parallel to DB, while CA is also equal and parallel to EG, therefore DB is also equal and parallel to EG. [XI. 9] And the straight lines DE, BG join their extremities; therefore DE is parallel to BG.

由CA平行且等于DB，且CA平行且等于EG，得DB平行且等于EG，连接DE、BG，则DE平行于BG，故角EDT等于角BGT，角DTU等于角GTS。

[I. 33] Therefore the angle EDT is equal to the angle BGT, for they are alternate; [I. 29] and the angle DTU is equal to the angle GTS. [I. 15] Therefore DTU, GTS are two triangles which have two angles equal to two angles, and one side equal to one side, namely that subtending one of the equal angles, that is, DU equal to GS, for they are the halves of DE, BG; therefore they will also have the remaining sides equal to the remaining sides. [I. 26] Therefore DT is equal to TG, and UT to TS.

在三角形DTU和GTS中，两角及一边（DU等于GS）相等，故全等，得DT等于TG，UT等于TS。