Question 50
In an acute triangle $ABC$ with $AB \lt AC$, the altitudes $BE$ and $CF$ intersect at $H$. The tangent to the circumcircle of $ABC$ at $A$ intersects the circumcircle of $AEF$ at $T \neq A$. The circumcircles of $TBE$ and $TCF$ intersect at $K \neq T$. Prove that $\angle KHB = \angle ABC$.