ISL 2023 G6

Let $A B C$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $B C$ at $D$. Let $A B$ and $A C$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $B C$ such that $B$ is the midpoint of $D M$ and $C$ is the midpoint of $D N$. Lines $M P$ and $N Q$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $K A$ meets the circumcircle of triangle $I J K$ at $X \neq K$. $$\text { Prove that } \angle B X P=\angle C X Q \text {. }$$