The two pairs of sides of an inscribed quadrilateral meet at $G$ & $H$; angle bisectors at $G$ & $H$ meet at $K$. Prove $\angle GKH$ is a right angle.

Question

Given that the two pairs of opposite sides of an inscribed quadrilateral meet when extended at $G$ and $H$ and the bisectors of the angles at $G$ and $H$ meet at K, prove that the $\angle GKH$ is a right angle.

Another proof that I'm completely stumped on. I'm thinking it has something to do with the angles of the quadrilateral and secants. Any hints/advice are greatly appreciated.

Answer 1

What OP expressed in the comment does lead to a solution.

Show by angle chasing that $2 \angle GKH = \angle GBH + \angle GDH = 180^\circ$.
Hence $ \angle GKH = 90^\circ$.

If you get stuck, explain how you've tried to demonstrate this equation.

Answer 2

Let the intersection of $GE$ and $KH$ be point $P$. Let $\angle B$ and $\angle E$ be $\beta$ and $\alpha$ respectively.

$\angle GPK$ and $\angle KGP$ can be found by simple angle chasing. Now just add them and show that it's $90^{\circ}$.