# Proof: By Euclid

• Let $A$, $B$, $C$, $D$, and $E$ have been conceived as the angular points of a pentagon having been inscribed (in circle $ABCDE$) [Prop. 3.11], such that the circumferences $AB$, $BC$, $CD$, $DE$, and $EA$ are equal.
• And let $GH$, $HK$, $KL$, $LM$, and $MG$ have been drawn through (points) $A$, $B$, $C$, $D$, and $E$ (respectively)1, touching the circle.
• And let the center $F$ of the circle $ABCDE$ have been found [Prop. 3.1].
• And let $FB$, $FK$, $FC$, $FL$, and $FD$ have been joined.
• And since the straight line $KL$ touches (circle) $ABCDE$ at $C$, and $FC$ has been joined from the center $F$ to the point of contact $C$, $FC$ is thus perpendicular to $KL$ [Prop. 3.18].
• Thus, each of the angles at $C$ is a right angle.
• So, for the same (reasons), the angles at $B$ and $D$ are also right angles.
• And since angle $FCK$ is a right angle, the (square) on $FK$ is thus equal to the (sum of the squares) on $FC$ and $CK$ [Prop. 1.47].
• So, for the same (reasons), the (square) on $FK$ is also equal to the (sum of the squares) on $FB$ and $BK$.
• So that the (sum of the squares) on $FC$ and $CK$ is equal to the (sum of the squares) on $FB$ and $BK$, of which the (square) on $FC$ is equal to the (square) on $FB$.
• Thus, the remaining (square) on $CK$ is equal to the remaining (square) on $BK$.
• Thus, $BK$ (is) equal to $CK$.
• And since $FB$ is equal to $FC$, and $FK$ (is) common, the two (straight lines) $BF$, $FK$ are equal to the two (straight lines) $CF$, $FK$.
• And the base $BK$ [is] equal to the base $CK$.
• Thus, angle $BFK$ is equal to [angle] $KFC$ [Prop. 1.8].
• And $BKF$ (is equal) to $FKC$ [Prop. 1.8].
• Thus, $BFC$ (is) double $KFC$, and $BKC$ (is double) $FKC$.
• So, for the same (reasons), $CFD$ is also double $CFL$, and $DLC$ (is also double) $FLC$.
• And since circumference $BC$ is equal to $CD$, angle $BFC$ is also equal to $CFD$ [Prop. 3.27].
• And $BFC$ is double $KFC$, and $DFC$ (is double) $LFC$.
• Thus, $KFC$ is also equal to $LFC$.
• And angle $FCK$ is also equal to $FCL$.
• So, $FKC$ and $FLC$ are two triangles having two angles equal to two angles, and one side equal to one side, (namely) their common (side) $FC$.
• Thus, they will also have the remaining sides equal to the (corresponding) remaining sides, and the remaining angle to the remaining angle [Prop. 1.26].
• Thus, the straight line $KC$ (is) equal to $CL$, and the angle $FKC$ to $FLC$.
• And since $KC$ is equal to $CL$, $KL$ (is) thus double $KC$.
• So, for the same (reasons), it can be shown that $HK$ (is) also double $BK$.
• And $BK$ is equal to $KC$.
• Thus, $HK$ is also equal to $KL$.
• So, similarly, each of $HG$, $GM$, and $ML$ can also be shown (to be) equal to each of $HK$ and $KL$.
• Thus, pentagon $GHKLM$ is equilateral.
• So I say that (it is) also equiangular.
• For since angle $FKC$ is equal to $FLC$, and $HKL$ was shown (to be) double $FKC$, and $KLM$ double $FLC$, $HKL$ is thus also equal to $KLM$.
• So, similarly, each of $KHG$, $HGM$, and $GML$ can also be shown (to be) equal to each of $HKL$ and $KLM$.
• Thus, the five angles $GHK$, $HKL$, $KLM$, $LMG$, and $MGH$ are equal to one another.
• Thus, the pentagon $GHKLM$ is equiangular.
• And it was also shown (to be) equilateral, and has been circumscribed about circle $ABCDE$.
• Thus, an equilateral and equiangular pentagon has been circumscribed about the given circle].
• (Which is) the very thing it was required to do.

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### References

1. Presumably, by finding the center of the circle [Prop. 3.1] ":https://www.bookofproofs.org/branches/finding-the-center-of-a-given-circle/, drawing straight lines between the center and points $A$, $B$, $C$, $D$, and $E$, and then drawing $GH$, $HK$, $KL$, $LM$, and $MG$ through thoses points, at right angles to the aforementioned straight line [Prop. 1.11] (translator's note).