Let $ABC$ be an arbitrary triangle. Let $D$ be an intersection of perpendicular bisector of side $AB$ and angle bisector of angle $ACB$ ($\angle ACD=\angle BCD$). Prove that quadrilateral ACBD is cyclic.
I can prove that triangle $ABD$ is isosceles. I have also figured out that $\angle DBA=\angle DCA$ and $\angle BDC=\angle BAC$ but I don't know how to prove that. I would like to avoid using analytic geometry. I don't necesarilly need full solution, any hint would be highly appreciated. Thank you in advance