17th Turkish Mathematical Olympiad

Since $\angle RPD=\angle RAD=\angle A^{\prime }AC=\angle A^{\prime }PC=\angle DPC$, $P,R,C$ are collinear. Then $\angle PRD=\angle PAD=\angle PAB=\angle PCB$ implies that $DR\parallel BC$. Similarly, $SE\parallel BC$, and consequently, $SE\parallel DR$ and $\frac{VD}{DA}=\frac{SR}{RA}$.

Let $l$ be tangent line through $A$ to the circumcircle of $ABC$, and let $T$ and $U$ be the points where $l$ intersects the lines $DS$ and $SE$, respectively. Also let $V$ be the point of intersection of $AB$ and $SE$.

We have $\angle UAE=\angle UAC=\angle B$, $\angle AEU=\angle AVE+\angle EAV=\angle B+\angle A$, and $\angle UAS=\angle UAE+\angle EAS=\angle AVS+\angle SAV=\angle ASU$. From these we conclude that $US=UA$ and then $\frac{US}{UE}=\frac{UA}{UE}=\frac{\sin \angle C}{\sin \angle B}=\frac{SV}{ES}$ where we used the law of sines in the triangle $AEU$ and the fact that $AS$ is the angle bisector of the triangle $VAE$.

$$\frac{AT}{TU}\cdot \frac{US}{SV}\cdot \frac{VD}{DA}=1\text{ by Menelaus’ Theorem for the triangle }AUV\text{ and the line }DS.$$ Substituting the last two ratios from above we obtain $$\frac{AT}{TU}\cdot \frac{UE}{ES}\cdot \frac{SR}{RA}=1$$ Applying Menelaus' Theorem for the triangle $AUS$ we conclude that the points $R,E,T$ are collinear.