, 2016

Solution:

The proof proceeds through a series of seven lemmas.

Lemma 1. Lines $DP$ and $EF$ are the internal and external angle bisectors of $\angle BPC$.

Proof. Since $DEF$ is the cevian triangle of $ABC$ with respect to its Gergonne point, we have that
$$-1=(\overline{EF} \cap \overline{BC}, D ; B, C)$$
Then since $\angle DPF=90^\circ$ we see $P$ is on the Apollonian circle of $BC$ through $D$. So the conclusion follows.

Lemma 2. Triangles $BPF$ and $CEP$ are similar.

Proof. Invoking the angle bisector theorem with the previous lemma gives
$$\frac{BP}{BF}=\frac{BP}{BD}=\frac{CP}{CD}=\frac{CP}{CE}$$
But $\angle BFP=\angle CEP$, so $\triangle BFP \sim \triangle CEP$.

Lemma 3. Quadrilateral $BZYC$ is cyclic; in particular, line $YZ$ is the antiparallel of line $BC$ through $\angle BAC$.

Proof. Remark that $\angle YBZ=\angle PBF=\angle ECP=\angle YCZ$.

Lemma 4. The circumcircles of triangles $AYZ$, $AEF$, $ABC$ are concurrent at a point $X$ such that $\triangle XBF \sim \triangle XCE$.

Proof. Note that line $EF$ is the angle bisector of $\angle BPZ=\angle CPY$. Thus
$$\frac{ZF}{FB}=\frac{ZP}{PB}=\frac{YP}{PC}=\frac{YE}{EC}$$
Then, if we let $X$ be the Miquel point of quadrilateral $Z Y C B$, it follows that the spiral similarity mapping segment $BZ$ to segment $CY$ maps $E$ to $F$; therefore the circumcircle of $\triangle AEF$ must pass through $X$ too.

Lemma 5. Ray $XP$ bisects $\angle FXE$.

Proof. The assertion amounts to
$$\frac{XF}{XE}=\frac{BF}{EC}=\frac{FP}{PE}$$
The first equality follows from the spiral similarity $\triangle BFX \sim \triangle CEX$, while the second is from $\triangle BFP \sim \triangle CEP$. So the proof is complete by the converse of angle bisector theorem.

Lemma 6. Points $X, P, I$ are collinear.

Proof. On one hand, $\angle FXI=\angle FAI=\frac{1}{2} \angle A$. On the other hand, $\angle FXP=\frac{1}{2} \angle FXE=\frac{1}{2} \angle A$. Hence, $X, P, I$ collinear.

Lemma 7. Points $X, Q, I$ are collinear.

Proof. On one hand, $\angle AXQ=90^\circ$, because we established earlier that line $YZ$ was antiparallel to line $BC$ through $\angle A$, hence $AQ \perp BC$ means exactly that $\angle AZQ=AYQ=90^\circ$. On the other hand, $\angle AXI=90^\circ$ according to the fact that $X$ lies on the circle with diameter $AI$. This completes the proof of the lemma.

Finally, combining the final two lemmas solves the problem.