Proof: By Euclid

• So let angles $BCD$ and $CDE$ have been cut in half by the (straight lines) $CF$ and $DF$, respectively [Prop. 1.9].
• And let the straight lines $FB$, $FA$, and $FE$ have been joined from point $F$, at which the straight lines meet, to the points $B$, $A$, and $E$ (respectively).
• So, similarly, to the (proposition) before this (one), it can be shown that angles $CBA$, $BAE$, and $AED$ have also been cut in half by the straight lines $FB$, $FA$, and $FE$, respectively.
• And since angle $BCD$ is equal to $CDE$, and $FCD$ is half of $BCD$, and $CDF$ half of $CDE$, $FCD$ is thus also equal to $FDC$.
• So that side $FC$ is also equal to side $FD$ [Prop. 1.6].
• So, similarly, it can be shown that $FB$, $FA$, and $FE$ are also each equal to each of $FC$ and $FD$.
• Thus, the five straight lines $FA$, $FB$, $FC$, $FD$, and $FE$ are equal to one another.
• Thus, the circle drawn with center $F$, and radius one of $FA$, $FB$, $FC$, $FD$, or $FE$, will also go through the remaining points, and will have been circumscribed.
• Let it have been (so) circumscribed, and let it be $ABCDE$.
• Thus, a circle has been circumscribed about the given pentagon, which is equilateral and equiangular.
• (Which is) the very thing it was required to do.

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