China National Team Selection Test

$$\begin{array}{rl}P{F}^2& =AF\cdot PF+PA\cdot PF\\ & =E{F}^2+PI\cdot PD.\end{array}\stackrel{◯}{1}$$ Since $PQ$ is the diameter of circle $O$ and point $I$ is on $AQ$, we see that $AI\perp AP$. Consequently, $$\angle IDF=\angle IAP=90^{\circ }.$$ Thus, we have $$P{F}^2-E{F}^2=P{D}^2-E{D}^2.$$ Combining ①, we have $$PI\cdot PD=P{D}^2-E{D}^2.$$ Thus $$Q{D}^2=E{D}^2=P{D}^2-PI\cdot PD=ID\cdot PD.$$ Consequently, we have $$△QID\sim △PQD.②$$ Since $PQ$ is the diameter of circle $O$, we see that $BP\perp BQ$. Suppose that $PQ$ is the perpendicular bisector of $BC$ at point $M$. Note that $I$ is the incentre of $△ABC$. We have $Q{I}^2=Q{B}^2=QM\cdot QP$. Thus, $$△QMI\sim △QIP.③$$ By ② and ③, we see that $\angle IQD=\angle QPD=\angle QPI=\angle QIM$. Hence, $MI\parallel QD$. Let $IK\perp BC$ be at $K$. Then $IK\parallel PM$; thus, $$\frac{PM}{IK}=\frac{PD}{ID}=\frac{PQ}{MQ}.$$ By the Circle-Power Theorem and the Sine Theorem, we know that $$\begin{array}{rl}PQ\cdot IK& =PM\cdot MQ=BM\cdot MC\\ & ={\left(\frac{1}{2}BC\right)}^2=(R\sin \angle BAC{)}^2,\end{array}$$ thus, ${\sin }^2\angle BAC=\frac{PQ\cdot IK}{R^2}=\frac{2R\cdot r}{R^2}=\frac{2r}{R}$. $\square$