Regular Polygon is composed of Isosceles Triangles

Theorem
Let $P$ be a regular $n$-gon.

Let $O$ be the center of $P$.

Then there exists a triangulation of $P$ into $n$ congruent isosceles triangles.

The three vertices of each triangle are $O$ and two adjacent vertices of $P$.

Proof
From Regular Polygon is Cyclic, it follows that $O$ is the center of the circumcircle of $P$.

Let $V_1, V_2$ be adjacent vertices of $P$.

By definition of circumcircle, $V_1 O$ and $V_2 O$ are radii of the circumcircle.

As $V_1 O$ and $V_2 O$ have the same length, it follows that $\triangle V_1 V_2 O$ is an isosceles triangles.

As there are $n$ pairs of adjacent vertices in $P$, corresponding to the $n$ sides of $P$, it follows that there are $n$ triangles of the type $\triangle V_1 V_2 O$.

All of these triangles consist of one side of $P$, as well as two sides of the type $V_1 O, V_2 O$ which they share with one of the adjacent triangles.

Hence, the triangles form a triangulation of $P$.

By definition of regular polygon, all sides of $P$ have the same length.

From Triangle Side-Side-Side Equality, it follows that all the triangles are congruent.