Problem

Source: Bulgarian TST 2020 P1

Tags: geometry, circumcircle



In acute triangle $\triangle ABC$, $BC>AC$, $\Gamma$ is its circumcircle, $D$ is a point on segment $AC$ and $E$ is the intersection of the circle with diameter $CD$ and $\Gamma$. $M$ is the midpoint of $AB$ and $CM$ meets $\Gamma$ again at $Q$. The tangents to $\Gamma$ at $A,B$ meet at $P$, and $H$ is the foot of perpendicular from $P$ to $BQ$. $K$ is a point on line $HQ$ such that $Q$ lies between $H$ and $K$. Prove that $\angle HKP=\angle ACE$ if and only if $\frac{KQ}{QH}=\frac{CD}{DA}$.