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.
(The Elements: Book $\text{XIII}$: Proposition $7$)
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 $\angle CBA = \angle BAE$.
Therefore from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:
- $AC = BE$
and:
- $\triangle ABC = \triangle ABE$
Thus:
- $\angle BCA = \angle BEA$
- $\angle ABE = \angle CAB$
So from Proposition $6$ of Book $\text{I} $: Triangle with Two Equal Angles is Isosceles:
- $AF = CF$
But we have that:
- $AC = BE$
Therefore:
- $FC = FE$
But:
- $CD = DE$
Therefore:
- $FC = FE$
and
- $CD = ED$
while $FD$ is common.
So from Proposition $8$ of Book $\text{I} $: Triangle Side-Side-Side Congruence:
- $\angle FCD = \angle FED$
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.
Therefore from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:
- $BE = BD$
and:
- $\triangle ABE = \triangle BCD$
Therefore:
- $\angle AEB = \angle CDB$
But $BE = BD$.
Therefore from Proposition $5$ of Book $\text{I} $: Isosceles Triangle has Two Equal Angles:
- $\angle BED = \angle BDE$
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 proof is Proposition $7$ of Book $\text{XIII}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XIII}$. Propositions