Question 59
Let $ABC$ be a right triangle with $\angle ACB=90^\circ$ and $CB\gt CA$. Choose $Q$ on the semicircle of diameter $CB$ so that $CQ=AC$, and choose $P$ on segment $CB$ so that $BP=BQ$. Let $S$ be the intersection of line $AP$ with the semicircle. Prove that $\angle CBS=\angle QBS$. The semicircle and $A$ lie on the same side of $BC$.