Finding center of convex polygon

Question

If I'm given vertices of a convex polygon (in the attached image, they are D,E,F,G and H) if we know that inside the polygon there exists a point (say O) for which each angle created by any two adjacent two vertices and the O are equal. That means, angle DOE, angle DOH, angle HOG, angle GOF and angle FOE all are equal. How to find O?

Answer 1

For a given $n$-gon $DEF\ldots$ and given that such $O$ exist, then all of the angles $\angle DOE=\angle EOF=\ldots = \frac{2\pi}n$ can be found. Denote the double of that angle as $\theta = 2\angle DOE$.

For any two neighbouring edges, say $DE$ and $EF$, draw an isosceles triangle for each edge with the edge as the base side, and $\theta$ as the opposite angle. The new isosceles triangle should lie inside the $n$-gon. Call the two new opposite vertices $A$ and $B$ respectively.

At $A$ with radius $AE$, and at $B$ with radius $BE$, draw two circles. The two circles intersect at $O$ and at $E$.