# Proof: By Euclid

• For let angles $BCD$ and $CDE$ have each been cut in half by each of the straight lines $CF$ and $DF$ (respectively) [Prop. 1.9].
• And from the point $F$, at which the straight lines $CF$ and $DF$ meet one another, let the straight lines $FB$, $FA$, and $FE$ have been joined.
• And since $BC$ is equal to $CD$, and $CF$ (is) common, the two (straight lines) $BC$, $CF$ are equal to the two (straight lines) $DC$, $CF$.
• And angle $BCF$ [is] equal to angle $DCF$.
• Thus, the base $BF$ is equal to the base $DF$, and triangle $BCF$ is equal to triangle $DCF$, and the remaining angles will be equal to the (corresponding) remaining angles which the equal sides subtend [Prop. 1.4].
• Thus, angle $CBF$ (is) equal to $CDF$.
• And since $CDE$ is double $CDF$, and $CDE$ (is) equal to $ABC$, and $CDF$ to $CBF$, $CBA$ is thus also double $CBF$.
• Thus, angle $ABF$ is equal to $FBC$.
• Thus, angle $ABC$ has been cut in half by the straight line $BF$.
• So, similarly, it can be shown that $BAE$ and $AED$ have been cut in half by the straight lines $FA$ and $FE$, respectively.
• So let $FG$, $FH$, $FK$, $FL$, and $FM$ have been drawn from point $F$, perpendicular to the straight lines $AB$, $BC$, $CD$, $DE$, and $EA$ (respectively) [Prop. 1.12].
• And since angle $HCF$ is equal to $KCF$, and the right angle $FHC$ is also equal to the [right angle] $FKC$, $FHC$ and $FKC$ are two triangles having two angles equal to two angles, and one side equal to one side, (namely) their common (side) $FC$, subtending one of the equal angles.
• Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26].
• Thus, the perpendicular $FH$ (is) equal to the perpendicular $FK$.
• So, similarly, it can be shown that $FL$, $FM$, and $FG$ are each equal to each of $FH$ and $FK$.
• Thus, the five straight lines $FG$, $FH$, $FK$, $FL$, and $FM$ are equal to one another.
• Thus, the circle drawn with center $F$, and radius one of $G$, $H$, $K$, $L$, or $M$, will also go through the remaining points, and will touch the straight lines $AB$, $BC$, $CD$, $DE$, and $EA$, on account of the angles at points $G$, $H$, $K$, $L$, and $M$ being right angles.
• For if it does not touch them, but cuts them, it follows that a (straight line) drawn at right angles to the diameter of the circle, from its extremity, falls inside the circle.
• The very thing was shown (to be) absurd [Prop. 3.16].
• Thus, the circle drawn with center $F$, and radius one of $G$, $H$, $K$, $L$, or $M$, does not cut the straight lines $AB$, $BC$, $CD$, $DE$, or $EA$.
• Thus, it will touch them.
• Let it have been drawn, like $GHKLM$ (in the figure).
• Thus, a circle has been inscribed in the given pentagon which is equilateral and equiangular.
• (Which is) the very thing it was required to do.

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