Question 155
A point $D$ is chosen on the leg $AC$ of an isosceles triangle $ABC$ with base $BC$. A point $K$ is chosen on the smaller arc $CD$ of the circumcircle of the triangle $BCD$. The ray $CK$ meets the line through $A$ parallel to $BC$ at $T$. Let $M$ be the midpoint of $DT$. Prove that $\angle AKT = \angle CAM$.