South African Mathematics Olympiad Third Round

Since $\angle AEI=\angle AFI=90^{\circ }$, the points $A$, $E$, $F$, $I$ lie on a circle with diameter $AI$. Moreover, since $\angle PFE=\angle BCA$ and $\angle PEF=\angle ABC$ by our assumptions on $P$, triangles $ABC$ and $PEF$ are similar, so $\angle EPF=\angle BAC=\angle EAF$. $P$ was assumed to lie on the same side of $EF$ as $A$, thus it follows that $P$ also lies on the same circle as $A$, $E$, $F$ and $I$. We also know that $BDIF$ is a cyclic quadrilateral (using the same reasoning as before, namely that $\angle BDI=\angle BFI=90^{\circ }$), so $\angle FID+\angle FBD=180^{\circ }$. Now we can conclude that $\angle PIF=\angle PEF=\angle ABC=\angle FBD=180^{\circ }-\angle FID$, so $\angle PIF+\angle FID=180^{\circ }$, which means that $P$, $I$, $D$ lie on a straight line.