Question 73
Let $ABCD$ is a trapezoid with $\angle A = \angle B = 90^{\circ}$ and let $E$ is a point lying on side $CD$. Let the circle $\omega$ is inscribed to triangle $ABE$ and tangents sides $AB$, $AE$ and $BE$ at points $P$, $F$ and $K$ respectively. Let $KF$ intersects segments $BC$ and $AD$ at points $M$ and $N$ respectively, as well as $PM$ and $PN$ intersect $\omega$ at points $H$ and $T$ respectively. Prove that $PH = PT$.