# Proof: By Euclid

• Let two planes of the aforementioned cube [Prop. 13.15], $ABCD$ and $CBEF$, (which are) at right angles to one another, be laid out.
• And let the sides $AB$, $BC$, $CD$, $DA$, $EF$, $EB$, and $FC$ have each been cut in half at points $G$, $H$, $K$, $L$, $M$, $N$, and $O$ (respectively).
• And let $GK$, $HL$, $MH$, and $NO$ have been joined.
• And let $NP$, $PO$, and $HQ$ have each been cut in extreme and mean ratio at points $R$, $S$, and $T$ (respectively).
• And let their greater pieces be $RP$, $PS$, and $TQ$ (respectively).
• And let $RU$, $SV$, and $TW$ have been set up on the exterior side of the cube, at points $R$, $S$, and $T$ (respectively), at right angles to the planes of the cube.
• And let them be made equal to $RP$, $PS$, and $TQ$.
• And let $UB$, $BW$, $WC$, $CV$, and $VU$ have been joined.
• I say that the pentagon $UBWCV$ is equilateral, and in one plane, and, further, equiangular.

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