Question 94
Let $O$ be a point inside the parallelogram $A B C D$ such that
$$
\angle A O B + \angle C O D = \angle B O C + \angle A O D
$$
Prove that there exists a circle $k$ tangent to the circumscribed circles of the triangles $\triangle A O B$, $\triangle B O C$, $\triangle C O D$ and $\triangle D O A$.
$$
\angle A O B + \angle C O D = \angle B O C + \angle A O D
$$
Prove that there exists a circle $k$ tangent to the circumscribed circles of the triangles $\triangle A O B$, $\triangle B O C$, $\triangle C O D$ and $\triangle D O A$.