Denote by $\triangle A_1B_1C_1$ medial triangle of $\triangle ABC$ and $\triangle PB_HC_H$ it's orthic($B_H\in AC$...). After inversion WRT circle centered at $A$ with radius $\sqrt{AO\cdot AP}$ composed with reflection over angle bisector we get that original problem is equivalent to proving that if $AO\cap (APB_1)=\{ E'\}$ and $AO\cap (APC_1)=\{ E'\}$, then $C_1E'$ and $B_1D'$ concur on $A-$median, we'll prove infact that they concur at $A_1$. \begin{align*} \angle D'B_1C&=\angle D'PA\\ &=\underbrace{180^{\circ}-\underbrace{\angle PD'A}_{=\angle PB_1B_H}}_{\text{Note NPC}}-\underbrace{\angle PAD'}_{=\angle PAO}\\ &=\angle PC_HB_H-(\angle BAC-2\angle BAP)\\ &= \cancel{\angle BAC}+\cancel{\angle ABC}-\angle BCA-\cancel{\angle BAC}+180^{\circ}-\cancel{2}\angle CBA\\ &= \angle BAC\\ &\Longrightarrow B_1D'||AB\\ &\Longrightarrow B_1D'\cap C_1E'=\{ A_1\}.\ \blacksquare \end{align*}

$AF $ is symmedian indeed by simple angle chase we find that $FA$ is the angle bisector of $\angle BFC=2\angle A$ RH HAS

We can prove that $\angle CEA = \angle BDA = \pi - \alpha$. Using this we can use that $B, F, O, C$ are cyclic. If we let $X$ be the intersection of the tangents to $\odot ABC$ at $B, C$, then $X$ lies on this circle too. Now an easy angle-chase gives us that $A-F-X$ are collinear. and hence $F$ lies on the symmedian of $\triangle ABC$. As $P$ is the midpoint of $BC$, the angle bisectors of $\angle FAP$ and $\angle BAC$ coincide.

[asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.953807897581248, xmax = 12.055914609922647, ymin = -8.210190520695098, ymax = 7.246406871059211; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); /* draw figures */ draw(circle((-3.127351049997505,-0.7225203498328047), 6.673530466256668), linewidth(0.4)); draw((-3.127351049997505,-0.7225203498328053)--(-3.17,-3.58), linewidth(0.4)); draw(circle((-9.672320371676197,1.6767446520133293), 5.188288429974637), linewidth(0.4) + dtsfsf); draw(circle((0.5723862591124261,2.9148249288269876), 6.970874835398997), linewidth(0.4) + dtsfsf); draw((-5.212599459966822,-0.9745241427848881)--(2.86,-3.67), linewidth(0.4)); draw((-5.212599459966822,-0.9745241427848881)--(-6.0412716085805265,0.7119573477177431), linewidth(0.4)); draw((-3.17,-3.58)--(-5.972583062566267,5.314089930673123), linewidth(0.4)); draw((-9.2,-3.49)--(-5.212599459966822,-0.9745241427848881), linewidth(0.4)); draw((-9.2,-3.49)--(-5.972583062566267,5.314089930673123), linewidth(0.8) + rvwvcq); draw((-5.972583062566267,5.314089930673123)--(2.86,-3.67), linewidth(0.8) + rvwvcq); draw((2.86,-3.67)--(-9.2,-3.49), linewidth(0.8) + rvwvcq); draw((-6.104676924992221,-3.536198851865788)--(-5.972583062566267,5.314089930673123), linewidth(0.4)); draw((2.86,-3.67)--(-6.0412716085805265,0.7119573477177431), linewidth(0.4) + linetype("4 4")); draw((-9.2,-3.49)--(-3.127351049997505,-0.7225203498328053), linewidth(0.4) + linetype("4 4")); /* dots and labels */ dot((-5.972583062566267,5.314089930673123),dotstyle); label("$A$", (-5.895057809190697,5.5195318521183845), NE * labelscalefactor); dot((-9.2,-3.49),dotstyle); label("$B$", (-9.625960627890027,-3.9889404743953008), NE * labelscalefactor); dot((2.86,-3.67),dotstyle); label("$C$", (2.9951506216528476,-4.2234543658564005), NE * labelscalefactor); dot((-6.104676924992221,-3.536198851865788),linewidth(4pt) + dotstyle); label("$A_1$", (-6.236168924043207,-4.116857142464991), NE * labelscalefactor); dot((-3.127351049997505,-0.7225203498328053),linewidth(4pt) + dotstyle); label("$O$", (-3.038252222300925,-0.5565098811919302), NE * labelscalefactor); dot((-3.17,-3.58),linewidth(4pt) + dotstyle); label("$P$", (-3.1022105563357707,-4.116857142464991), NE * labelscalefactor); dot((-6.082784341398924,-2.06939575111493),linewidth(4pt) + dotstyle); label("$D$", (-5.788460585799288,-2.3899821235241654), NE * labelscalefactor); dot((-6.0412716085805265,0.7119573477177431),linewidth(4pt) + dotstyle); label("$E$", (-6.577280038895718,0.530781797400442), NE * labelscalefactor); dot((-5.212599459966822,-0.9745241427848881),linewidth(4pt) + dotstyle); label("$F$", (-4.850405019954885,-0.9189404407227209), NE * labelscalefactor); dot((-6.057880962571258,-0.40086936966126746),linewidth(4pt) + dotstyle); label("$H$", (-6.577280038895718,-0.49255154715708477), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Since, $\odot (ADB)$,$\odot (AEC)$ are tangent to $AC$,$AB$ at $A$, $\implies$ $\Delta ABF \sim \Delta CAF$ $\implies$ $F$ is the midpoint of $A-$symmedian chord

Somehow this felt as isogonal sister of 2017G3 Solution 1 Claerly $\angle{ADB}=\angle AEC=180^{\circ}-\angle A$.Therefore $\angle BFC=360^{\circ}-\angle AFB-\angle AFC=2\angle A \implies B,F,O,C \text{concyclic}$.If $X=BB\cap CC \text{w.r.t} (ABC)$.Its well known that $B,X,O,C$ concylic.Therefore $\angle BFX=\angle BCX=\angle A=180^{\circ}-AFB \implies A,F,X$ collinear $\implies$ $AF$ symmedian.Hence the conclusion.$\square$ Solution 2 This is trivial if one knows The dumpty points.Similar to above solution since $\angle ADB=180^{\circ}-\angle A \implies (ADB)$ tangent to $AC$ and similarly $(AEC)$ tagent to $AB \implies$ $F$ is the dumpty point so done.$\square$

Here is my solution for this problem Solution Let $S$ be intersection of tangents at $B$, $C$ of ($ABC$) $SB$, $SC$ intersect ($ABD$), ($ACE$) again at $T$, $U$ We have: $\widehat{TBD}$ = $\widehat{TBA}$ + $\widehat{ABD}$ = $\widehat{ACB}$ + $\widehat{ABO}$ = $90^o$ But: $A$, $D$, $B$, $T$ lie on a circle then: $\widehat{TAD}$ = $90^o$ or $TA$ $\perp$ $AD$ Similarly: $UA$ $\perp$ $AD$ So: $T$, $A$, $U$ are collinear Hence: $\widehat{TBC}$ + $\widehat{TUC}$ = $\widehat{TBA}$ + $\widehat{ABC}$ + $\widehat{BCS}$ = $\widehat{ACB}$ + $\widehat{ABC}$ + $\widehat{BAC}$ = $180^o$ or $T$, $B$, $C$, $U$ lie on a circle But: $TU$ $\parallel$ $BC$ then: $TBCU$ is isosceles trapezoid So: $P_{S / (ABD)}$ = $\overline{SB}$ . $\overline{ST}$ = $\overline{SC}$ . $\overline{SU}$ = $P_{S / (ACE)}$ or $S$ lies on radical axis of ($ABD$), ($ACE$) Hence: $A$, $F$, $S$ are collinear or $AF$ is $A$ - symmedian of $\triangle$ $ABC$ Therefore: internal angle bisector of $\widehat{FAP}$ passes through incenter of $\triangle$ $ABC$

Here's another Inversive Solution... Macedonia MO 2017 wrote: Let $O$ be the circumcenter of the acute triangle $ABC$ ($AB < AC$). Let $A_1$ and $P$ be the feet of the perpendicular lines drawn from $A$ and $O$ to $BC$, respectively. The lines $BO$ and $CO$ intersect $AA_1$ in $D$ and $E$, respectively. Let $F$ be the second intersection point of $\odot ABD$ and $\odot ACE$. Prove that the angle bisector of $\angle FAP$ passes through the incenter of $\triangle ABC$. Solution: Let $\odot(ADB)=\omega_1$ and $\odot(ACE)=\omega_2$. Now we will perform a $\sqrt{\frac{bc}{2}}$ Inversion ($\Psi$) centered at $A$ with a reflection along the angle bisector of $\angle BAC$. Let $X,Y$ be the midpoints of $AB,AC$. So, $$\begin{cases} \Psi: B\leftrightarrow Y \\ D=AA_1\cap BO\leftrightarrow AO\cap \odot(AYA_1)=D'\end{cases} \implies \Psi:\omega_1\leftrightarrow YD'$$ Similarly we will map $\omega_2$... $$\begin{cases} \Psi:C\leftrightarrow X \\ \Psi: E=AA_1\cap CO\leftrightarrow AO\cap \odot(AXA_1)=E'\end{cases}\implies \Psi:\omega_2\leftrightarrow XE'$$ So, $$\Psi: F=\omega_1\cap \omega_2 \leftrightarrow YD'\cap XE'$$. Now we will work on this problem... ... wrote: If $ABC$ is a triangle and $AA_1$ be the feet of perpendicular from $A$ to $BC$. Let $X$ be the midpoint of $AB$. Let $O$ be the circumcenter of $\triangle ABC$. $P$ be the midpoint of $BC$ and let $AO\cap\odot(AXA_1)=E'$. Then $\overline{X-P-E'}$ $$\angle BAA_1=\angle XA_1A=\angle XE'A=\angle CAE'\implies XE'\|AC$$also $XP\|AC$. Hence, $\overline{X-P-E'}$. Similarly we get that $\overline{Y-D'-P}$. So, $XE'\cap YD'=P$. Hence, $$\Psi:F=\omega_1\cap\omega_2\leftrightarrow XE'\cap YD'=P$$Hence, $AF,AP$ are isogonal. Precisely $AF$ is the $A-\text{Symmedian}$ of $\triangle ABC$. So, angle bisector of $\angle FAP$ is same as the angle bisector of $\angle BAC$. So it must pass through the incenter of $\triangle ABC$. $\blacksquare$

Stefan4024 wrote: Let $O$ be the circumcenter of the acute triangle $ABC$ ($AB < AC$). Let $A_1$ and $P$ be the feet of the perpendicular lines drawn from $A$ and $O$ to $BC$, respectively. The lines $BO$ and $CO$ intersect $AA_1$ in $D$ and $E$, respectively. Let $F$ be the second intersection point of $\odot ABD$ and $\odot ACE$. Prove that the angle bisector of $\angle FAP$ passes through the incenter of $\triangle ABC$. Nice problem! We claim that $F$ is indeed the $A$-Dumpty Point of $\triangle ABC$, which implies that $\overline{AF}$ is the $A$-Symmedian and $\overline{AP}$ is the $A$-median which are known to be isogonal to each other with respect to $\angle BAC$. Also this solution is based on the diagram on page 90 of ABJTOG pls see that diagram. We let $Y$ and $Z$ be the points where $\odot (ABD)$ and $\odot (ACE)$ cut $\overline{BC}$ again (same notation as ABJTOG) We see that $\angle BFC = \angle BFD + \angle DFZ + \angle ZFC = \angle BAD + \angle DBA + \angle ZCA + \angle ZAC = (180 - \angle BDA) + (180 - \angle AZC) = (180 - \angle BDA) + (180 - \angle AEC) = \angle OED + \angle ODE = \angle BOC$ implying that points $B, F, O, C$ are concyclic. Now $\angle FCA = \angle OCA + \angle OCF = 90 - \angle ABC + \angle OBF = \angle BAD + \angle DAF = \angle BAF$, similarly $\angle FBA = \angle CAF$ which means that $F$ is indeed $A$-Dumpty Point of $\triangle ABC$, as desired.

It suffices to show that $\angle FAB = \angle PAC$. Because $\angle BAD = \angle ACE$ and $\angle OED = \angle ODE$ by a simple angle chase, $\triangle ADB \sim \triangle CEA$. By the spiral similarity lemma, $F$ is the spiral center of this pair of triangles, so correspondingly $\triangle AFB \sim \triangle CFA$. Now let $\angle PAC =y $ and $\angle BAF =x$. We have $$\frac{\sin(A-y)}{\sin y} = \frac{AC}{AB}$$and $$\frac{\sin(A-x)}{\sin x} = \frac{AF}{BF} = \frac{AC}{AB}$$as $\angle ABD = \angle A - x$ by the Ratio Lemma and Law of Sines, respectively. The corresponding function for $x$ is decreasing for $\angle A$ acute and $x < A$, so we must have $x=y$.