Question 159
Let $ABC$ be an acute-angled triangle. Let $D$ and $E$ be points on $\overline{BC}$ and $\overline{AC}$, respectively, such that $\overline{AD} \perp \overline{BC}$ and $\overline{BE} \perp \overline{AC}$. Let $P$ be the point where $\overrightarrow{AD}$ meets the semicircle constructed outwardly on $\overline{BC}$, and $Q$ the point where $\overrightarrow{BE}$ meets the semicircle constructed outwardly on $\overline{AC}$. Prove that $PC = QC$.