By Miquel's theorem, $BDPE$ is cyclic. So power of point on $A$ gives $BCEF$ is cyclic. By Reim's theorem, tangents at $A$ of $\odot(APEF)$ is parallel to $BC$ so circumcenter $O$ of $\Delta AEF$ lies on altitude from $A$ to $BC$. Furthermore, $$A(B, C ; D, \infty)\implies (E, F; P, A)=-1$$hence $\frac{EQ}{FQ}=\frac{EP}{FP}=\frac{EA}{FA}$ so $\odot(APQ)$ is $A$-Apollonius circle of $\Delta AEF$. But it's well known that this circle is orthogonal to $\odot(AEF)$. Let $O$ be the center of $\odot(AEF)$, we see that $AO$ is tangent to $\odot(APQ)$ but $AO\perp BC$ so we are done.

Back to the main problem, By Miquel and Power of Point, we get $D,P,E,C$ and $B,F,E,C$ concyclic. So $\angle C=\angle APE=\angle AFE$ so $BC$ is parallel to the tangent in A at $(AEF)$ so they intersect at the point at infinity and $AO$ will be perpendicular on $BC$. Since $(B, C ; D, \infty)=-1$ we get that $AFPE$ is harmonic. By the lemma, we are done.

cirey wrote: In a non-isosceles acute triangle $ABC$, $D$ is the midpoint of the edge $[BC]$. The points $E$ and $F$ lie on $[AC]$ and $[AB]$, respectively, and the circumcircles of $CDE$ and $AEF$ intersect in $P$ on $[AD]$. The angle bisector from $P$ in triangle $EFP$ intersects $EF$ in $Q$. Prove that the tangent line to the circumcirle of $AQP$ at $A$ is perpendicular to $BC$. Claim: $\overline{AQ}$ bisects angle $BAC$. (Proof) Observe $\angle AEF=\angle APF=\angle ABC$ and $\angle AFE=\angle APE=\angle ACB$. Hence $$\frac{FA}{EA}=\frac{\sin B}{\sin C}=\frac{\sin DAB}{\sin DAC}=\frac{\sin PEF}{\sin PFE}=\frac{FP}{EP}=\frac{FQ}{QE}$$proving the claim. $\blacksquare$ Now let $\overline{AL}$ be the $A$-altitude. Then $\angle LAQ=\angle APQ=\tfrac{1}{2}(\angle B-\angle C)$ proving the result.

Ax//BC>xAE=ACD=APE>Ax tangent (AEF)>-1=A(BCDx)=A(FPEA)>FPEA harmonic>AQ bisect A>QAH=1/2(B-C)=APQ

Let $H$ be the foot of the altitude from $A$ to $BC$. We will show that $AH$ is tangent to the circumcircle of $AQP$. $\angle ECD=\angle APE=\angle AFE$, thus $BCEF$ is cyclic. Also, $EF$ and $CB$ are antiparallel with respect to the angle $BAC$. $AD$ is the median in the triangle $ABC$. Then, $AP$ is the symmedian in the triangle $AEF$. Also $\angle FPQ=\angle QPE$. Hence, $\angle FAQ=\angle QAE$. Let $\angle QPA=\alpha$ and $\angle APE=\beta$. Then, $\angle BCE=\angle AFE=\angle APE=\beta, \angle FBC=\angle AEF=\angle APF=2\alpha+\beta$. So $\angle BAQ=\frac{\angle BAC}{2}=90-\alpha-\beta$. In conclusion, $\angle HAQ=\angle BAQ-\angle BAH=(90-\alpha-\beta)-(90-2\alpha-\beta)=\alpha =\angle QPA$. Done.

Let the foot from $A$ to $BC$ be $G$. Let $AG\cap (APEF) = J$. We have $\angle AJF = \angle APF = \angle ABC = 90 - \angle FAJ$. So the circumcenter of $(APEF)$ lies on $AG$. So, $ -1 = (B,C;D,BC_{\infty}) \stackrel{A} = (F,E;P,A) \implies$ $AQ$ is the angle bisector of $\angle FAE$. Now a straightforward angle chase does the job: $\angle APQ = \angle APF - \angle QPF = \angle ABC - (90 - \frac{1}{2}\angle A) = \frac{1}{2}\angle A + \angle B - 90 = \angle QAG$. $\blacksquare$

Nearly, same solution with Barış Koyuncu. $\angle ACD = \angle APE = \angle AFE$. Therefore, $BCEF$ is a cyclic quadrilateral. $\angle ABC = \angle AEF = \angle FPA$. Therefore, $BFPD$ is also a cyclic quadrilateral. If we write down the similarities (\(\triangle AFP \sim \triangle ADB\) and \(\triangle AEP \sim \triangle ADC\)), we obtain: \[ \frac{AF}{AD} = \frac{FP}{DB} \quad \text{and} \quad \frac{AE}{AD} = \frac{PE}{CD} \tag{1} \] By cross-multiplying the equalities: \[ \frac{AF}{AE} = \frac{FP}{PE} \tag{2} \] Due to the Angle Bisector Theorem, \(\frac{FP}{PE} = \frac{FQ}{QE}\), so \(\frac{AF}{AE} = \frac{FQ}{QE}\), implying that \(AQ\) is the angle bisector of \(\angle FAE\). Let the line tangent to the circumcircle of \(\triangle AQP\) at point \(A\) intersect \(BC\) at \(H\). We are asked to show that \(AH\) is the altitude of \(\triangle ABC\). From the tangent-chord angle, \(\angle PAQ = \angle QAH\). If \(\angle ABC = \angle FPA = \beta\) and \(\angle ACB = \angle APE = \theta\), then \(\angle FPA = |\theta - \beta|\). \[ \angle FAQ = \angle QAE = \frac{180^\circ - 2\beta - 2\theta}{2} = 90^\circ - \beta - \theta \] If \(\angle C > \angle B\), then \(\angle HAE = \angle QAE - \angle QAH = 90^\circ - \beta - \theta - (\theta - \beta) = 90^\circ - 2\theta = 90^\circ - \angle ACB\), hence \(AH \perp BC\). If \(\angle B > \angle C\), then \(\angle HAE = \angle QAE + \angle QAH = 90^\circ - \beta - \theta + (\beta - \theta) = 90^\circ - 2\theta = 90^\circ - \angle ACB\), and again \(AH \perp BC\).

Define the angle bisector of $\angle BAC$ as $l$, and define $X=l\cap EF$, and $M=l\cap (ABC)$, $M\neq A$. Define, moreover, the foot of the perpendicular from $A$ as $Y$. Angle chasing: $\measuredangle PDB=\measuredangle CDP = \measuredangle PEC=\measuredangle=\measuredangle AEP=\measuredangle PFA=\measuredangle PFB$. $\Longrightarrow$ $F, B, D, P$ is cyclic. $\newline$ Power of a point at $A$ gives: $AE\cdot AC= AP\cdot AD = AF\cdot AB$ $\newline$ $\Longrightarrow B, F,E,C$ is cyclic. $\measuredangle BFX=\measuredangle BFE=\measuredangle C = \measuredangle AMB = \measuredangle XMB$ $\newline$ $\Longrightarrow B, F, X, M$ and symmetrically $C, E, X, M$ are cyclic. $\newline$ By power of a point from $A$ again, we get: $\newline$ $AX\cdot AM=AF\cdot AB=AP\cdot AD$ $\newline$ $\Longrightarrow X, M, P, D$ is cyclic. $\newline$ Angle chasing gives: $\newline$ We know $\angle MXE = 180 - \angle C + \angle \frac{A}{2}$, also $\angle MXP = 180 - (\angle PDB + 90) = 90 - (180 - \angle CDP) = \angle CDP - 90$. Moreover, that $\angle XEP = 180 - \angle B - (180 - \angle CDP) = \angle CDP - \angle B$ $\newline$ Combining: $\newline$ $\angle XPE = 180 - \angle PXE - \angle XEP = 180 - (\angle MXE - \angle MXP) - \angle XEP = 180 - (180 - \angle C - \angle \frac{A}{2} - \angle CDP - 90) - \angle CDP - \angle B = 90 - \angle \frac{A}{2}$. $\newline$ $\Longrightarrow$ Thus, $X$ is on the angle bisector of $\angle EPF$ $\newline$ Finishing with an angle chase: $\angle XPA = \angle XPE - \angle EPA = 90 - \angle \frac{A}{2} - \angle C$. $\newline$ Simultaneously, $\angle YAX = \angle BAX - \angle BAY = \angle \frac{A}{2} - (90 - \angle B) = \angle B + \angle \frac{A}{2} - 90 = 180 - \angle A - \angle C + \angle \frac{A}{2} - 90 = 90 - \angle \frac{A}{2} - \angle C$ $\newline$ $\Longrightarrow$ Thus, $\angle YAX = \angle XAP$, and $AY$ is tangent to $(AXP)$, this finishes the problem. $\blacksquare$