17th Junior Turkish Mathematical Olympiad

Let $\angle EAP=\alpha$ and $\angle ECD=\beta$. Note that $\angle RPE=\alpha$. Choose a point $X$ on $[ED]$ such that $\angle RPX=90^{\circ }$. Observe that $R,P,X,E$ are cyclic and hence $\angle RXE=\alpha$ which implies that $A,R,X,D$ are concyclic. Therefore $ER\cdot EA=EX\cdot ED$. On the other hand we have $ER\cdot EA=E{P}^2=E{C}^2$. Thus we get $EX\cdot ED=E{C}^2$ which implies that $\angle EXC=\beta$. Clearly $\angle QDC=180^{\circ }-\alpha -\beta$ and hence $\angle QBC=\angle RBC=\alpha +\beta$ since $Q,B,C,D$ are cyclic and $B,Q,R$ are collinear. On the other hand $\angle RXC=\alpha +\beta$ as well and therefore $X$ is the reflection of $B$ with respect to $E$. Thus we get $\angle EBC=\angle EXC=\beta$ and the result follows.