NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Consider the mirror image $E$ of point $C$ with respect to $A$. Since $\angle CDB=\angle CEB=45^{\circ }$, the quadrilateral $BCDE$ is cyclic, and $CE$ is a diameter of its circumcircle. Hence $A$ is its circumcenter, implying $AB=AD$.