If the sides of the opposite planes of a cube are bisected, and the planes are carried through the points of section, then the intersection of the planes and the diameter of the cube bisect one another.

Let the sides of the opposite planes *CF* and *AH* of the cube *AF* be bisected at the points *K, L, M, N, O, Q, P,* and *R,* and through the points of section let the planes *KN* and *OR* be carried. Let *US* be the common section of the planes, and *DG* the diameter of the cube *AF.*

I say that *UT* equals *TS,* and *DT* equals *TG.*

Join *DU, UE, BS,* and *SG.*

Then, since *DO* is parallel to *PE,* therefore the alternate angles *DOU* and *UPE* equal one another.

Since *DO* equals *PE,* and *OU* equals *UP,* and they contain equal angles, therefore the base *DU* equals the base *UE,* the triangle *DOU* equals the triangle *PUE,* and the remaining angles equal the remaining angles. Therefore the angle *OUD* equals the angle *PUE.*

Therefore *DUE* is a straight line. For the same reason *BSG* is also a straight line, and *BS* equals *SG.*

Now, since *CA* equals and is parallel to *DB,* while *CA* also equals and is parallel to *EG,* therefore *DB* equals and is parallel to *EG.*

And the straight lines *DE* and *BG* join their ends, therefore *DE* is parallel to *BG.*

Therefore the angle *EDT* equals the angle *BGT,* for they are alternate, and the angle *DTU* equals the angle *GTS.*

Therefore *DTU* and *GTS* are two triangles which have two angles equal to two angles and one side equal to one side, namely that opposite one of the equal angles, that is, *DU* equals *GS,* for they are the halves of *DE* and *BG,* therefore the remaining sides equal the remaining sides. Therefore *DT* equals *TG,* and *UT* equals *TS.*

Therefore, *if the sides of the opposite planes of a cube are bisected, and the planes are carried through the points of section, then the intersection of the planes and the diameter of the cube bisect one another.*

Q.E.D.

There are a couple of details missing from this proof. For instance, it is not shown that *KL* and *MN* actually lie in one plane, and that the line *SU* actually intersects the line *DG.*