Japan Mathematical Olympiad

Let $F,G$ be the intersection points of lines $QD,QE$ and triangle $ABC$ other than $Q$ respectively. Let line $BG$ and $CF$ meet at $R$. Then, application of Pascal's theorem for six points $A,B,G,Q,F,C$ which are concyclic shows that $D,E,R$ are colinear. These six points are on the same circle in the order $A,F,B,Q,C,G$. In particular, $R$ lies on segment $DE$. We will show that $P=R$.

First, we show that $BP:PC=BR:RC$. Applying the sine rule to triangle $BRC$ provides that $$BR:RC=\sin \angle RCB:\sin \angle CBR=\sin \angle DQB:\sin \angle CQE.$$ On the other hand, applying the sine rule to triangle $DQB$ and triangle $CQE$ provides that $$BP:PC=QE:QD=CE\cdot \frac{\sin \angle ECQ}{\sin \angle CQE}:BD\cdot \frac{\sin \angle QBD}{\sin \angle DQB}$$ From $\angle ECQ+\angle QBD=180^{\circ }$ and $BD=CE$ we obtain $$BP:PC=\sin \angle DQB:\sin \angle CQE=BR:RC.$$ Assume that $P$ and $R$ are distinct. In case that $D,P,R,E$ lie in that order, we have $$\begin{gathered} \angle CBR<\angle CBP<\angle CBA<90^{\circ }, \\ \angle PCB<\angle RCB<\angle ACB<90^{\circ }. \end{gathered}$$ This shows that $\sin \angle CBR<\sin \angle CBP$, $\sin \angle PCB<\sin \angle RCB$. By the sine rule we obtain $$\frac{PC}{BP}=\frac{\sin \angle CBP}{\sin \angle PCB}>\frac{\sin \angle CBR}{\sin \angle RCB}=\frac{RC}{BR}$$ which contradicts $BP:PC=BR:RC$. In case that $D,R,P,E$ lie in that order, we have a contradiction similarly. Thus we have $P=R$ by contradiction.

From above we have $$\angle BPC=\angle BRC=180^{\circ }-\angle RCB-\angle CBR=180^{\circ }-\angle DQB-\angle CQE.$$ In addition, we have $$\angle DQB+\angle CQE=\angle CQB-\angle EQD=180^{\circ }-\angle BAC-\angle EQD.$$ Putting them together, we obtain $\angle BPC=\angle BAC+\angle EQD$, which yields the desired conclusion.