Question 55
The triangle $ABC$ is given. Point $T$ is the intersection of the $A$-symmedian and the circumcircle of $ABC$. Point $D \neq A$ lies on the line $AC$ in such a way that $BD = BA$. The line tangent to the circumcircle of $ADT$ at point $D$ intersects the circumcircle of $DCT$, for the second time at point $K$. Prove that $\angle BKC = 90^\circ$.