You are given a right triangle $ABC$ with $\angle ACB = 90^\circ$. Let $W_A , W_B$ respectively be the midpoints of the smaller arcs $BC$ and $AC$ of the circumcircle of $\triangle ABC$, and $N_A , N_B$ respectively be the midpoints of the larger arcs $BC$ and $AC$ of this circle. Denote by $P$ and $Q$ the points of intersection of segment $AB$ with the lines $N_AW_B$ and $N_BW_A$, respectively. Prove that $AP = BQ$. Proposed by Oleksiy Masalitin