# Proof: By Euclid

• For let three angles of the equilateral pentagon $ABCDE$ - first of all, the consecutive (angles) at $A$, $B$, and $C$ - -be equal to one another.
• I say that pentagon $ABCDE$ is equiangular.

• For let $AC$, $BE$, and $FD$ have been joined.
• And since the two (straight lines) $CB$ and $BA$ are equal to the two (straight lines) $BA$ and $AE$, respectively, and angle $CBA$ is equal to angle $BAE$, base $AC$ is thus equal to base $BE$, and triangle $ABC$ equal to triangle $ABE$, and the remaining angles will be equal to the remaining angles which the equal sides subtend [Prop. 1.4], (that is), $BCA$ (equal) to $BEA$, and $ABE$ to $CAB$.
• And hence side $AF$ is also equal to side $BF$ [Prop. 1.6].
• And the whole of $AC$ was also shown (to be) equal to the whole of $BE$.
• Thus, the remainder $FC$ is also equal to the remainder $FE$.
• And $CD$ is also equal to $DE$.
• So, the two (straight lines) $FC$ and $CD$ are equal to the two (straight lines) $FE$ and $ED$ (respectively).
• And $FD$ is their common base.
• Thus, angle $FCD$ is equal to angle $FED$ [Prop. 1.8].
• And $BCA$ was also shown (to be) equal to $AEB$.
• And thus the whole of $BCD$ (is) equal to the whole of $AED$.
• But, (angle) $BCD$ was assumed (to be) equal to the angles at $A$ and $B$.
• Thus, (angle) $AED$ is also equal to the angles at $A$ and $B$.
• So, similarly, we can show that angle $CDE$ is also equal to the angles at $A$, $B$, $C$.
• Thus, pentagon $ABCDE$ is equiangular.
• And so let consecutive angles not be equal, but let the (angles) at points $A$, $C$, and $D$ be equal.
• I say that pentagon $ABCDE$ is also equiangular in this case.
• For let $BD$ have been joined.
• And since the two (straight lines) $BA$ and $AE$ are equal to the (straight lines) $BC$ and $CD$, and they contain equal angles, base $BE$ is thus equal to base $BD$, and triangle $ABE$ is equal to triangle $BCD$, and the remaining angles will be equal to the remaining angles which the equal sides subtend [Prop. 1.4].
• Thus, angle $AEB$ is equal to (angle) $CDB$.
• And angle $BED$ is also equal to (angle) $BDE$, since side $BE$ is also equal to side $BD$ [Prop. 1.5].
• Thus, the whole angle $AED$ is also equal to the whole (angle) $CDE$.
• But, (angle) $CDE$ was assumed (to be) equal to the angles at $A$ and $C$.
• Thus, angle $AED$ is also equal to the (angles) at $A$ and $C$.
• So, for the same (reasons), (angle) $ABC$ is also equal to the angles at $A$, $C$, and $D$.
• Thus, pentagon $ABCDE$ is equiangular.
• (Which is) the very thing it was required to show.

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