Equilateral Pentagon is Equiangular if Three Angles are Equal

Proof

 * Euclid-XIII-7.png

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 $\angle CBA = \angle BAE$.

Therefore from :
 * $AC = BE$

and:
 * $\triangle ABC = \triangle ABE$

Thus:
 * $\angle BCA = \angle BEA$
 * $\angle ABE = \angle CAB$

So from :
 * $AF = CF$

But we have that:
 * $AC = BE$

Therefore:
 * $FC = FE$

But:
 * $CD = DE$

Therefore:
 * $FC = FE$

and
 * $CD = ED$

while $FD$ is common.

So from :
 * $\angle FCD = \angle FED$

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.

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.

Therefore from :
 * $BE = BD$

and:
 * $\triangle ABE = \triangle BCD$

Therefore:
 * $\angle AEB = \angle CDB$

But $BE = BD$.

Therefore from :
 * $\angle BED = \angle BDE$

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.