# Proof: By Euclid

• For let the opposite planes $CF$ and $AH$ of the cube $AF$ have been cut in half at the points $K$, $L$, $M$, $N$, $O$, $Q$, $P$, and $R$.
• And let the planes $KN$ and $OR$ have been produced through the pieces.
• And let $US$ be the common section of the planes, and $DG$ the diameter of cube $AF$.
• I say that $UT$ is equal to $TS$, and $DT$ to $TG$.

• For let $DU$, $UE$, $BS$, and $SG$ have been joined.
• And since $DO$ is parallel to $PE$, the alternate angles $DOU$ and $UPE$ are equal to one another [Prop. 1.29].
• And since $DO$ is equal to $PE$, and $OU$ to $UP$, and they contain equal angles, base $DU$ is thus equal to base $UE$, and triangle $DOU$ is equal to triangle $PUE$, and the remaining angles (are) equal to the remaining angles [Prop. 1.4].
• Thus, angle $OUD$ (is) equal to angle $PUE$.
• So, for this (reason), $DUE$ is a straight line [Prop. 1.14].
• So, for the same (reason), $BSG$ is also a straight line, and $BS$ equal to $SG$.
• And since $CA$ is equal and parallel to $DB$, but $CA$ is also equal and parallel to $EG$, $DB$ is thus also equal and parallel to $EG$ [Prop. 11.9].
• And the straight lines $DE$ and $BG$ join them.
• $DE$ is thus parallel to $BG$ [Prop. 1.33].
• Thus, angle $EDT$ (is) equal to $BGT$.
• For (they are) alternate [Prop. 1.29].
• And (angle) $DTU$ (is equal) to $GTS$ [Prop. 1.15].
• So, $DTU$ and $GTS$ are two triangles having two angles equal to two angles, and one side equal to one side - (namely), that subtended by one of the equal angles - (that is), $DU$ (equal) to $GS$.
• For they are halves of $DE$ and $BG$ (respectively).
• (Thus), they will also have the remaining sides equal to the remaining sides [Prop. 1.26].
• Thus, $DT$ (is) equal to $TG$, and $UT$ to $TS$.
• Thus, if the sides of the opposite planes of a cube are cut in half, and planes are produced through the pieces, then the common section of the (latter) planes and the diameter of the cube cut one another in half.
• (Which is) the very thing it was required to show.

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