Problem

Source: Iran MO Third Round G1

Tags: geometry, Angle Chasing, Spiral Similarity, circumcircle, geometric transformation, reflection, angle bisector



Triangle $ABC$ is assumed. The point $T$ is the second intersection of the symmedian of vertex $A$ with the circumcircle of the triangle $ABC$ and the point $D \neq A$ lies on the line $AC$ such that $BA=BD$. The line that at $D$ tangents to the circumcircle of the triangle $ADT$, intersects the circumcircle of the triangle $DCT$ for the second time at $K$. Prove that $\angle BKC = 90^{\circ}$(The symmedian of the vertex $A$, is the reflection of the median of the vertex $A$ through the angle bisector of this vertex).