Common Section of Bisecting Planes of Cube Bisect and are Bisected by Diagonal of Cube

Proof

 * Euclid-XI-38.png

Let $AF$ be a cube.

Let the edges of the opposite faces $CF$ and $AH$ of $AF$ be bisected at the points $K, L, M, N, O, P, Q, R$.

Let the plane $KN$ be constructed through $K, L, M, N$.

Let the plane $OR$ be constructed through $O, P, Q, R$.

Let $US$ be the common section of $KN$ and $OR$.

Let $DG$ be the diameter of the cube $AF$.

It is to be demonstrated that:
 * $UT = TS$
 * $DT = TG$

Let $DU, UE, BS, SG$ be joined.

We have that $DO$ is parallel to $PE$.

Therefore from :
 * $\angle DOU = \angle UPE$

We have that:
 * $DO = PE$
 * $OU = UP$
 * $\angle DOU = \angle UPE$

So from :
 * $\triangle DOU = \triangle PUE$

Therefore:
 * $\angle OUD = \angle PUE$

So from :
 * $DUE$ is a straight line.

For the same reason:
 * $BSG$ is a straight line.

and:
 * $BS = SG$

We have that:
 * $CA = DB$ and $CA \parallel DB$

and:
 * $CA = EG$ and $CA \parallel EG$

Therefore from:
 * $DB = EG$ and $DB \parallel EG$

We have that the straight lines $DE$ and $BG$ join the endpoints of $DB$ and $EG$.

Therefore from :
 * $DE \parallel BG$

Therefore from :
 * $\angle EDT = \angle BGT$

Therefore from :
 * $\angle DTU = \angle GTS$

Therefore from :
 * $\triangle DTU = \triangle GTS$

Therefore:
 * $DT = TG$

and
 * $UT = TS$