# Equilateral Pentagon is Equiangular if Three Angles are Equal

## Theorem

In the words of Euclid:

If three angles of an equilateral pentagon, taken either in order or not in order, be equal, the pentagon will be equiangular.

## Proof Let $ABCDE$ be an equilateral pentagon.

Let $A, B, C$ be three vertices of $ABCDE$ taken in order such that $\angle A = \angle B = \angle C$.

Let $AC, BE, FD$ be joined.

We have that $CB$ and $BA$ are equal to $BA$ and $AE$ respectively.

We also have that $\angle CBA = \angle BAE$.

$AC = BE$

and:

$\triangle ABC = \triangle ABE$

Thus:

$\angle BCA = \angle BEA$
$\angle ABE = \angle CAB$
$AF = CF$

But we have that:

$AC = BE$

Therefore:

$FC = FE$

But:

$CD = DE$

Therefore:

$FC = FE$

and

$CD = ED$

while $FD$ is common.

$\angle FCD = \angle FED$

But we have already proved that:

$\angle BCA = \angle AEB$

Therefore:

$\angle BCD = \angle AED$

But by hypothesis:

$\angle BCD = \angle A = \angle B$

Therefore:

$\angle AED = \angle A = \angle B$

Similarly it can be proved that:

$\angle CDE = \angle A = \angle B = \angle C$

Therefore $ABCDE$ is equiangular.

$\Box$

Next, suppose that $A, C, D$ are the three vertices of $ABCDE$ such that $\angle A = \angle C = \angle D$.

Let $BD$ be joined.

We have that $BA = BC$ and $AE = CD$.

Also, they contain equal angles.

$BE = BD$

and:

$\triangle ABE = \triangle BCD$

Therefore:

$\angle AEB = \angle CDB$

But $BE = BD$.

$\angle BED = \angle BDE$

Therefore:

$\angle AED = \angle DDE$

But by hypothesis:

$\angle CDE = \angle A = \angle C$

Therefore:

$\angle AED = \angle A = \angle C$

For the same reason:

$\angle ABC = \angle A = \angle C = \angle D$

Therefore $ABCDE$ is equiangular.

$\blacksquare$

## Historical Note

This theorem is Proposition $7$ of Book $\text{XIII}$ of Euclid's The Elements.