Let $ABCD$ be a circumscribed quadrilateral and let $X$ be the intersection point of its diagonals $AC$ and $BD$. Let $I_1, I_2, I_3, I_4$ be the incenters of $\triangle DXC$, $\triangle BXC$, $\triangle AXB$, and $\triangle DXA$, respectively. The circumcircle of $\triangle CI_1I_2$ intersects the sides $CB$ and $CD$ at points $P$ and $Q$, respectively. The circumcircle of $\triangle AI_3I_4$ intersects the sides $AB$ and $AD$ at points $M$ and $N$, respectively. Prove that $AM+CQ=AN+CP$