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
$$\angle CBR < \angle CBP < \angle CBA < 90^\circ, \\ \angle PCB < \angle RCB < \angle ACB < 90^\circ.$$
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.