Invert the diagram about $A$. In the new diagram, $B,C$ are midpoints of $AD,AE$, $(DBC)$ meets $DE$ at $Q$, and $(EBC)$ meets $DE$ at $P$. Note that $DBCQ$ is an isoscleles trapezoid, so $Q$ is the reflection of $D$ across $\ell$, the perpendicular bisector of $BC$. Similarly, $P$ is the reflection of $E$ across $\ell$. But the midpoint $M$ of $DE$ gets reflected to the foot of the altitude from $A$ to $DE$, so the midpoint of $PQ$ is the foot of the altitude from $A$ to $PQ$. Thus $AP=AQ$.

Nice! For reference, here's a diagram. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.47323965649598, xmax = 20.94485338438140, ymin = -10.17125405549053, ymax = 11.47108751470510; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((2.180000000000002,5.680000000000006)--(0.1400000000000002,0.9800000000000011)--(7.500000000000008,0.8000000000000009)--cycle, zzttqq); /* draw figures */ draw((2.180000000000002,5.680000000000006)--(0.1400000000000002,0.9800000000000011), zzttqq); draw((0.1400000000000002,0.9800000000000011)--(7.500000000000008,0.8000000000000009), zzttqq); draw((7.500000000000008,0.8000000000000009)--(2.180000000000002,5.680000000000006), zzttqq); draw(circle((3.015553673997120,3.920972447882108), 1.947390015184418)); draw(circle((3.817315899677342,0.7802501201400499), 3.682737058292900)); draw(circle((3.784021831163189,-0.5811073479942349), 3.964334907390584)); draw((4.927325926588292,4.291721319036744)--(-7.329950976601768,1.162689018449501)); draw((-7.329950976601768,1.162689018449501)--(4.840000000000005,3.240000000000003)); draw((0.1400000000000002,0.9800000000000011)--(-7.329950976601768,1.162689018449501)); draw(circle((-3.490372135120598,5.348459394818307), 5.680056278210315)); draw(circle((0.01152850651306523,-2.023825480604612), 8.003199088719365)); /* dots and labels */ dot((2.180000000000002,5.680000000000006),dotstyle); label("$A$", (2.322373428631665,5.898723821382306), NE * labelscalefactor); dot((0.1400000000000002,0.9800000000000011),dotstyle); label("$C$", (0.2731816188290878,1.189177732186912), NE * labelscalefactor); dot((7.500000000000008,0.8000000000000009),dotstyle); label("$B$", (7.643081987417304,1.009424064660370), NE * labelscalefactor); dot((1.160000000000002,3.330000000000004),dotstyle); label("$E$", (1.315752890483031,3.561926143537263), NE * labelscalefactor); dot((4.840000000000005,3.240000000000003),dotstyle); label("$D$", (4.982727708024484,3.454073943021338), NE * labelscalefactor); dot((1.520301899852382,2.673354716701205),dotstyle); label("$Q$", (1.675260225536114,2.878862206936404), NE * labelscalefactor); dot((4.927325926588292,4.291721319036744),dotstyle); label("$P$", (5.054629175035100,4.496645214675280), NE * labelscalefactor); dot((-7.329950976601768,1.162689018449501),dotstyle); label("$J$", (-7.168620216769744,1.368931399713454), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $BP\cap \odot ADE$ at $R$ Then $\angle{ARP}=180^{\circ}-\angle{AEP}=180^{\circ}-\angle{RBC}$so that $AR\parallel BC\parallel DE\implies ARDE$ is an isoceles trapezium. Let $CQ\cap \odot ADE$ at $R'$, similarly $AR'\parallel BC\implies R=R'$. Let $RF\perp BC$ at $F$, $AA'\perp BC$ at $A'$, $DD'\perp BC$ at $D'$, $EE'\perp BC$ at $E'$. Since $AA'\parallel EE'$ and $E$ is the midpoint of $AC$, $A'E'=E'C$. $ARFA'$ and $EDD'E'$ are rectangles$\implies D'F=A'E'$. Thus $FC=FA'+2A'E'=FA'+D'F+A'E'=DE=\frac{1}{2} BC\implies RB=RC$. Now $\angle{APQ}=\angle{ADQ}=\angle{RCB}=\angle{RBC}=\angle{AEP}=\angle{AQP}\implies AP=AQ$

Fixed; thanks. Doesn't really matter though, the problem doesn't change if you interchange $B$ and $C.$

it is easy by using complex number. let A=2a, B=2x, C=-2x then P=a^2/x +2a-x

First use Pascal to prove that $BP$ meets $CQ$ at circle $(ADE)$ at point $T$. Then let $M$ be the midpoint of $BC$.Let $k$ be the reflection of $M$ in $DE$.Angle chasing gives $k$ = $T$. we are done now.(Angle chase)

My solution: By angle-chasing we obtain: $\angle EPA=\angle DPB=\angle CBA=\angle EDA$ and $\angle EQC=\angle APD=180-\angle ACB=180-\angle DEA$ $\Longrightarrow $ By law of sines in $\triangle APB$ we get: $sin\angle PAB.sin\angle DPB=sin\angle PBA.sin\angle DPA...(1)$ since $PD$ is median $\Longrightarrow $ By law of sines in $\triangle AQC$ we get: $sin\angle CAQ.sin\angle EQC=sin\angle QCA.sin\angle EQA...(2)$ since $QE$ is median By $...(1)$ and $...(2)$ we get $sin\angle CAQ.sin\angle PAB=sin\angle QCA.sin\angle PBA$ $\Longrightarrow $ it is easy to see that: $\triangle PAB \sim \triangle QCA$ $\Longrightarrow $ $\angle QEC=\angle PDA$ as $P,D,Q,E,A$ are concyclic we get: $\angle QEC=\angle PDA=\angle QDA=\angle QPA=\angle PAQ$ $\Longrightarrow $ $\angle QPA=\angle PAQ$ then $AP=AQ$...

My solution: Let $\Phi$ be a composition of the symetry wtr bisector of $\angle BAC$ and inversion with the center $A$ and radius $\sqrt{AE\cdot AF}$. Let $P^*=\Phi(P)$. Notice that $AD=2AB^*, \ AE=2AC^*$. Furthermore points $P^*, \ D, \ E, \ Q^*$ are collinear so if quadrilaterals $B^*C^*P^*D$ and $B^*C^*EQ^*$ are cyclic, they must be isosceles trapezoids. Hence $Q^*B^*=C^*E=C^*A$ and $AB^*=B^*D=C^*P^*$. Due to that we have also $\angle Q^*B^*A=\pi - Q^*B^*D=\pi - \angle P^*C^*E=\angle P^*C^*A$, thus $\triangle Q^*B^*A \equiv \triangle P^*C^*A$ and we are done.

Using Radical Axes Theorem on $\odot (ADE), \odot (ABC) , \odot (BDC)$ and $\odot (ADE) , \odot (BDC) , \odot (BEC)$ $\implies$ $PD$ $ \cap$ $ QE $ $\cap$ $ BC$ $=$ $F$ and $AF$ tangent to $\odot (ABC)$ at $A$, now, $\angle AEQ$ $=$ $\angle QBC$ $=$ $\angle FAQ$ $\implies$ $FAQB$ cyclic and $\angle APD$ $=$ $\angle AED$ $=$ $\angle ACB$ $\implies$ $FAPC$ cyclic. Now, observe, $\angle APE$ $=$ $180^{\circ}-\angle ABC$ $=$ $\angle FAB$ $+$ $\angle BAC$ $=$ $\angle FAC$ $=$ $\angle FPC$ $\implies$ $PF$ is symmedian in $\Delta APC$ $\implies$ $FE$ is symmedian in $\Delta ABF$ $\implies$ $\Delta ADF$ $ \stackrel{+}{\sim}$ $ \Delta QBF$ $\implies$ $\Delta FQA$ $ \stackrel{+}{\sim} $ $\Delta FBD$ $\implies$ $\angle FAQ$ $=$ $\angle FDB$ $=$ $\angle ADP$ $=$ $\angle AQP$ $\implies$ $AF||PQ$ $\implies$ $AQ=AP$

Radax gives that $EP,DQ,BC$ are concurrent at , say $X$.Now $$\angle QXC = \angle EDQ = \angle QAC \implies ( AXCQ)$$.Also, $$ \angle PXB = \angle PED = \angle PAB \implies ( APBX)$$.Then $$ \angle XAC= \angle XQC = \angle B$$. Now $$ \angle APX = \angle ADE = \angle BPD = \angle B$$which means that $PX$ is a symmedian of $APB$.Since $X\in (APB)$ we have that $XD$ is the $X$ symmedian of $XAB$.THUS $$\angle AXD = \angle BXP \implies \angle PDA = \angle PEA = \angle B+ \angle AXP = \angle B+ \angle DXB = \angle ADQ \implies AP=AQ$$$400$ th post !

Claim1 : ∠EQA = ∠DPB and ∠CQE = ∠APD. Proof : ∠DPB = 180 - ∠ECB - ∠DPE = 180 - ∠ACB - ∠BAC = ∠ABC = ∠ADE = ∠AQE and ∠APD = ∠ADE + ∠EAD = 180 - ∠DQC + 180 - ∠DQE = ∠CQE. Claim2 : AQC and BPA are similar. Proof : AQ/CQ = AE/CE . Sin ∠CQE/Sin ∠EQA = Sin ∠CQE/Sin ∠EQA = Sin ∠APD/Sin ∠DPB = Sin ∠APD/Sin ∠DPB . AD/BD = BP/AP. from last part we also know that ∠AQC = ∠APB. ∠AQP = ∠AED - ∠PED = ∠ACB - ∠PAD = ∠ACB - ∠ACQ = ∠QCB = ∠ADQ = ∠APQ so AP = AQ. we're Done.

Let the tangent to $\odot(ABC)$ at $A$ meet $BC$ at $T$. By radical center on $\odot(ABC)$, $\odot(ADE)$, $\odot(BCDP)$, we get that $T\in DP$. Similarly, $T\in EQ$. Now, let $O\in\odot(ADE)$ be the circumcenter of $\triangle ABC$, and let $K$ be the midpoint of symmedian chord, so $OK\perp AT$. Then, observe that $(AK;DE)=-1$, so projecting from $T$ to the same circle gives $(AO;PQ) = -1$. However, since $AO$ is diameter, it follows that $AP=AQ$.