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.

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, 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.

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.