37th Iranian Mathematical Olympiad

First we prove that it suffices to show that $ASCBR$ is cyclic. Let $\omega$ be the circumcircle of $△ABC$. The fact $△PAE\sim △BDC$ implies that $\angle PAB=\angle C$. Therefore, $PA$ is tangent to $\omega$. Similarly, $△QAD\sim △EBC$ implies $\angle QAD=\angle B$. Therefore, $QA$ is tangent to $\omega$. Hence, points $A,P$ and $Q$ are collinear and $PQ$ is tangent to the $\omega$ at $A$.

Now we have $RS\parallel PQ$ and since $R$ and $S$ lie on $\omega$, we find out that $AR=AS$ and $\angle ARS=\angle ASR$. Let $M$ be the intersection point of lines $SD$ and $RE$. We have $$\angle MRA=\angle MSA=\angle BAO⟹\angle MRS+\angle ARS=\angle MSR+\angle ASR.$$

And since $\angle ARS=\angle ASR$, we have $\angle MRS=\angle MSR$ and $MS=MR$. Therefore $A,M$ and $O$ lie on the perpendicular bisector of $RS$.

Now let $X$ be the intersection point of lines $PE,BD$. Since $△PAE\sim △BDC$, angles $\angle APE$ and $\angle BDC$ are equal and $APXD$ is cyclic and since $R,X$ lie on the circumcircle of $△PAD$, it follows that $APRXD$ is cyclic. Furthermore, $$\angle XBE=\angle XRA-\angle ARE=\angle XDC-\angle BAC=\angle ABD=\angle XBE.$$ Therefore, $XBRE$ is cyclic as well. We'll get $$\begin{array}{rl}\angle BRE& =\angle EXD=180^{\circ }-(\angle C+\angle A)=\angle B\\ \Rightarrow \angle ARB& =\angle ARE+\angle BRE=\angle A+\angle B\\ \Rightarrow \angle ARB+\angle C& =180^{\circ }.\end{array}$$ Which gives us that $R$ lies on $\omega$. Analogously, $S$ lies on $\omega$. Therefore $ASCBR$ is cyclic and we're done. ■