This is ISL 2022 G2.

suppose $DP\cap \odot(ABD)=\{D,U\},EP\cap \odot(AEC)=\{E,V\},\odot(ABD)\cap \odot(ACE)=\{A,Z\}$ $\angle UAV=\angle UAB+\angle BAC+\angle CAV=180^{\circ}$ Which means that $U-V-A$ So $U,E,D,V$ are cyclic $\Rightarrow PD\cdot PU=PZ\cdot PV$ Hence $A-P-Z$ Let $AZ\cap XB=T$ Notice $\angle ABD=TBC$ Which means that points $A$ and $T$ are symmetric about $BC$ So $\angle ACB=\angle BCT$ Which means that $T-Y-C$ So $TB\cdot TX=TZ\cdot TA=TY\cdot TC$ Hence $X,B,Y,C$ are cyclic

$WLOG$ let $AB>AC$. Let $\omega_{1}=(ABD)$ and $\omega_{2}=(AEC)$. Claim 1: $AF\equiv \rho(\omega_{1}, \omega_{2})$ Proof: It is enough to show that $deg(F,\omega_{1}) = deg(F,\omega_{2})$ $\Leftrightarrow FD\cdot FB = FE\cdot FC$ $\Leftrightarrow \frac{FE}{FB} = \frac{FD}{FC}$. However, $\frac{FE}{FB} =\frac{FP}{FA}$ because $EP||AB$ and $\frac{FD}{FC}=\frac{FP}{FA}$ because $DP||CA$. Hence $\frac{FE}{FB} = \frac{FD}{FC} = \frac{FP}{FA}$. Now let $K$ be the symmetrical point of $A$ wrt to $BC$ $\Rightarrow K\in \rho(\omega_{1}, \omega_{2})$. Claim 2: $K\in BX$ and $K\in CY$ Proof: From $\angle DBA = \angle DXA = \angle DAX = \angle DBX = \beta$ and from $\angle KBF = \angle ABF = \beta$, we get that $\angle KBF = \angle XBF$. So $K\in BX$. Similarly, we get that $K\in CY$. Now $BXCY$ is cyclic $\Leftrightarrow KX\cdot KB = KC\cdot KY \Leftrightarrow deg(K,\omega{1}) = deg(K,\omega{2})$, but this is true since $K\in \rho(\omega_{1}, \omega_{2})$.

From Thales we have $\frac{FB}{FE} = \frac{FA}{FP} = \frac{FC}{FD}$, so $FB \cdot FD = FE \cdot FC$, i.e. $F$ lies on the radical axis of the circles of $ABD$ and $ACE$. So if $T$ is the second common point of these circles, then $A$, $F$, $T$ are collinear. Now note that if it turns out that $BX$, $CY$ and $AFT$ are concurrent, then from Power of a Point $XBYC$ would be cyclic. To argue the latter concurrency, let $BX \cap AFT = U$. With $\angle ABC = \beta$ we have $\angle ABX = \angle ADX = 180^{\circ} - 2\angle AXD = 180^{\circ} - 2\angle ABD = 180^{\circ} - 2\beta$ (relying on the given $DX = DA$), so $\angle UBF = 180^{\circ} - \angle ABC - \angle ABX = \beta = \angle ABC$. Therefore $ABU$ is isosceles and $BC$ is the perpendicular bisector of $AU$. Analogously if $CY \cap AFT = V$, then $BC$ is the perpendicular bisector of $AV$. Therefore $U\equiv V$ and we are done.

Nice problem, although it is trivial using geogebra Claim : $(ABD) \cap (ACE)$ lies on lines $AP$ Proof : Let $K = AP \cap (ABD)$ and $J$ is foot from $A$ to $BC$ it suffices to prove that $K \in (ACE)$ By power of point ; $AJ \times JK = BJ \times JD$ Since $\frac{AJ}{JP} = \frac{BJ}{JE}$ Divided first equation by second equation : $$JP \times JK= JE \times JD $$$\rightarrow K,E,P,D $ concylic then $\angle EKJ = \angle JDP = \angle JCA$ Thus $E$ lies on $(ACE)$ $\blacksquare$ Let $I = BX \cap CY$ By some angle chasing implies that $\triangle BIC \equiv \triangle BAC$ Then $I \in AP$ Which means $A-P-K-I$ By Claim implies $(ABD) \cap (ACE) \in AP$ Power of point : $ IB \times IX = IK \times IA = IY \times IC $ Hence $B,X,C,Y$ concyclic. $\blacksquare$

Nice problem! Let $XB \cap YC = T$. By angle chasing easy to show that $A, P, T$ are collinear. So, by PoP we need to prove that $A,K,P$ are collinear, where $K$ is second intersection of $(ABD),(ACE)$ or (by PoP) that $BH \cdot HD = CH \cdot HE$ where $H = AP \cap BC$, which is obvious because by Thales $\frac{HD}{HC}=\frac{HP}{HA}=\frac{HE}{HB}$

(1) Let F be the foot of perpendicular from $A$ to $BC$. Similarity yields $\frac{FC}{FD}=\frac{FA}{FP}=\frac{FB}{FE} => FB\times FD=FC\times FE$ (2) Let $G$ be the second intersection of $(ABD)$ and $(ACE)$ and let $F'=AG\cap BC$. By POP, $F'B \times F'D = F'C \times F'E$ From (1) and (2), we get $F'=F$. Hence, $F$ lies on the radical axis $AT$ of $(ABD)$ and $(ACE)$. Let $BX \cap CY=A'$ By simple angle chasing we get that $\angle ABC = \angle A'BC$ and $\angle ACB = \angle A'CB=>A'$ is the reflection of $A$ w.r.t. $BC$ and $A'$ lies on radical axis $AG$ which gives us $A'B \times A'X = A'Y \times A'C$. Hence, $B,X,C,Y$ are concyclic.

$AF \cap (ABD) = K \implies AF \cdot FK = BF \cdot FD = EF \cdot FC \implies A, E, K, C$ concyclic. Let $T (\neq A)$ be a point on $AK$ such that $AF = FT$. $DA = DX \implies \angle ABC = \angle XBC$, so $B, X, T$ are collinear. Similarly, $C, Y, T$ are collinear. $\therefore BX, AK, CY$ are concurrent at $T \implies TX \cdot TB = TK \cdot TA = TY \cdot TC \implies B, X, Y, C$ are concyclic.

Define by $A'$ the reflection of $A$ over line $BC$. Then observe that $\providecommand{\dang}{\measuredangle} \dang DBA = \dang DXA = \dang XAD = \dang XBD = \dang A'BD$, so $BX$ is the reflection of line $AB$ over $BC$. Hence $A'$, $B$, and $X$ are collinear. Similarly, $C$, $Y$, and $A'$ are collinear. Now observe that $\frac{EF}{FD} = \frac{BF}{FC}$ so $EF \cdot FC = BF \cdot FD$, implying that $AF$ is the radical axis of $(ABD)$ and $(ACE)$. Moreover, $A'$ lies on $AF$, so we obtain \[ A'X \cdot A'B = A'A \cdot A'P = A'Y \cdot A'C, \]and so $XBCY$ is cyclic as desired.

Great problem! Claim: The radical axis of $(ABD)$ and $(ACE)$ is line $AF$. Proof: From the parallel lines we have $\frac{FE}{FB}=\frac{FD}{FC}$ and rearranging we get $FE\cdot FC= FD\cdot FB$ as desired. Now the arc midpoints give that $\measuredangle ABC=\measuredangle AXD=\measuredangle DAX=\measuredangle CBX$ so $BC$ is the exterior bisector of $\angle XBA$ and similarly it’s the exterior bisector of $\angle ACY$. Then, let $A’$ be the flection of $A$ across $BC$. Then, $\angle A’BF=\angle CBA$ so $A’$ lies on $BX$ and similarly it lies on $CY$. The result follows by the radical lemma as $A’$ lies on $AF$

Using the parallel lines it is clear that $AF$ is the radical axis of $(ABD)$ and $(ACE)$. Let $XB \cap CY = Z$. Notice that by the arc midpoints conditions, BC is the angle bisector of $\angle ABZ$ and $\angle ACZ$. This implies $Z$ is the reflection of $A$ over $BC$, and therefore $Z$ lies on $AF$. This implies that $ZB \cdot ZX = ZY \cdot ZC$, which implies that $BXCY$ is cyclic.

Clearly $X,Y$ lie on external bisectors of angles $ABC,ACB$ respectively, so point $Z=BX\cap CY$ is a reflection of $A$ over $BC.$ By Thales $|EF|:|BF|=|FP|:|FA|=|FD|:|FC|\implies |EF|\cdot |FC|=|BF|\cdot |FD|,$ so $\overline{APFZ}$ is a radical axis of $\odot (ABD), \odot (ACE)$ implying the desired result by PoP.

Claim: $$EF\cdot CF=BF\cdot DF.$$Note that by by the parallel conditions, $$\frac{EF}{BF}=\frac{PF}{AF}=\frac{DF}{CF},$$which shows the claim. Thus, $F$ lies on the radical axis of $(ABD)$ and $(ACE)$, so the radical axis is $AF$. Claim 2: $XB,YC,AF$ concur at the reflection of $A$ across $F$. Let $Q_B$ be the intersection of $XB$ and $AF$. Note that by $AXBD$ cyclic and $DX=DA$, we have $$\angle ABD=\angle AXD=\angle XAD=\angle DBQ_B,$$$\triangle ABF\equiv \triangle Q_BBF$ so $Q_B$ is the reflection of $A$ across $F$. Similarly, the intersection of $AF$ and $CY$ is also the reflection of $A$ across $F$. Hence, $XB,YC,AF$. Thus, by Radical Axis Theorem, we are done since the radical axis of $(ABD)$ and $(ACE)$ is $AF$.

Solution using Angles and Radical Axis Since $$PE\parallel AB,PD\parallel AC,ED\parallel BC\Longrightarrow\triangle PED\sim\triangle ABC\Longrightarrow\frac{FE}{FB}=\frac{FD}{FC}\Longrightarrow FE\cdot FC=FD\cdot PB,$$which means that $F$ lies on the radical axis of $(ABD)$ and $(ACE)$. So, $AF$ is the radical axis of the two circles. Suppose $K\ne A$ is the second meeting point of the two circles. Then, $A,F,K$ are collinear. Suppose $XB$ and $YC$ are extended to meet at $L$. We claim that $L$ is the reflection of $A$ about $\overline{BC}$. Note that since $DX=XA$, we have $$\angle CBA=\angle DBA=\angle XBD=\angle LBD=\angle LBC$$and $\angle ACB=\angle BCL$. It follows that $L$ is indeed the reflection of $A$ about $\overline{BC}$. So, $F$ is the mid-point of $\overline{AL}$. Hence, $L$ lies on the radical axis of $(ABD)$ and $(ACE)$. So, we have $$LB\cdot LX=LC\cdot LY,$$which means that $B,C,X,Y$ are concyclic. $\square$

Why was this kinda easy Let $A'$ be the reflection of $A$ across $F.$ Claim: $AA'$ is the radical axis of $(ABD)$ and $(ACE).$ Proof: Let $(EPD)$ intersect $AF$ at $M.$ Then $$\measuredangle BDM = \measuredangle EDM = \measuredangle EPM = \measuredangle BAM$$from the parallel lines, so $ABDM$ is cyclic. Similarly, $ACEM$ is also cyclic. This means that $AM$ and hence $AA'$ is the radical axis of the two circles. Claim: $X$ lies on line $BA'.$ Proof: We have $$\measuredangle XBC = \measuredangle XBD = \measuredangle XAD = \measuredangle DXA = \measuredangle DBA = \measuredangle A'BC$$from $DX = DA$ and from the definition of $A'.$ Similarly, $Y$ lies on line $CA'.$ Hence $XB$ and $CY$ concur on the radical axis of $(ABD)$ and $(ACE),$ so the conclusion follows from the radical center lemma.

Moving Point Solution: Note that if we fix triangle ABC, then XBA equals 180-2ABC by angle chasing, which is clearly fixed, and similarly YCA is also fixed, meaning as P moves along AF, X and Y move along fixed lines. Also, note that XAD always has the same angles, and it is a well-known fact of moving point geometry that if XAD maintains the same internal angles, and A stays fixed, as D moves linearly along BC, X moves linearly along its own fixed-line. But clearly, as P moves linearly along AF, D moves linearly along BC, meaning as P moves linearly along AF, X moves linearly along its own fixed-line, and, symmetrically, Y does as well. Note the problem statement is equivalent to showing that XBC=XYC, but XBC is fixed, so we just need to show XYC stays equal to this fixed value. Thus, we only need to prove the claim for two different positions of P, as due to X and Y moving linearly together, any other position of P also has XYC at the same angle. So we use the following two cases: Case 1: A=P Then clearly the four points in question are all on the circumcircle of ABC Case 2: P=F Then clearly the four points in question are cyclic iff BX, AF, and CY concur. But it is easy to angle chase that BX and CY intersect at the reflection of A across BC, which is clearly on AF as desired. So the claim is proved.

Classical Solution: Claim: BX, AF, and CY concur. Proof: Let BX intersect with CY at G. Using brief angle chasing, we figure out that CBG=ABC and BCG=ACB, thus by ASA, ABC and GBC are congruent. Then, BGCA is a kite -> AG and BC are perpendicular -> F is on AG as desired. Now let H=(ABD) intersection with (ACE) but H isn't A. Claim: PEHD cyclic Proof: EPD=BAC=BAH+HAC=BDH+HEC=EDH+HED=180-DHE, claim proven. Claim: A, P, H collinear Proof: PHD=PED=ABD=AHD, proven. But now, GB*GX=GH*GA=GC*GY, done.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9 cm); 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 = -0.14935488767340993, xmax = 37.18654387787122, ymin = -12.284319427423974, ymax = 5.783807319442906; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); draw((5.897836584516442,2.062458721172221)--(4.037162285381103,-4.817139674315033)--(13.928115138679482,-3.837837411612221)--cycle, linewidth(1.) + zzttqq); /* draw figures */ draw((5.897836584516442,2.062458721172221)--(4.037162285381103,-4.817139674315033), linewidth(1.) + zzttqq); draw((4.037162285381103,-4.817139674315033)--(13.928115138679482,-3.837837411612221), linewidth(1.) + zzttqq); draw((13.928115138679482,-3.837837411612221)--(5.897836584516442,2.062458721172221), linewidth(1.) + zzttqq); draw(circle((9.255522291976863,-1.7824809254512008), 5.104665924060127), linewidth(1.)); draw(circle((7.213125537171397,-1.98469842592699), 4.255521824910829), linewidth(1.)); draw((5.076700794792333,-4.714215069422832)--(6.741818344632492,1.442337976696709), linewidth(1.)); draw((5.324966832003627,-0.055651811144906166)--(10.882860138479776,-4.139347807671598), linewidth(1.)); draw((3.0380516090263376,-2.8082418047576243)--(10.882860138479776,-4.139347807671598), linewidth(1.)); draw((10.882860138479776,-4.139347807671598)--(5.897836584516442,2.062458721172221), linewidth(1.)); label(".",(18.310492764274557,3.237621436415595),SE*labelscalefactor); draw((13.928115138679482,-3.837837411612221)--(7.21078230814106,-11.198293087436406), linewidth(1.)); draw((5.897836584516442,2.062458721172221)--(7.21078230814106,-11.198293087436406), linewidth(1.)); draw((3.0380516090263376,-2.8082418047576243)--(7.21078230814106,-11.198293087436406), linewidth(1.)); /* dots and labels */ dot((5.897836584516442,2.062458721172221),dotstyle); label("$A$", (5.995766810786723,2.3072842868479237), NE * labelscalefactor); dot((4.037162285381103,-4.817139674315033),dotstyle); label("$B$", (4.135092511651384,-4.57231410863933), NE * labelscalefactor); dot((13.928115138679482,-3.837837411612221),dotstyle); label("$C$", (14.026045364949763,-3.5930118459365183), NE * labelscalefactor); dot((6.1689485921196745,-0.6757725556204175),dotstyle); label("$P$", (6.265074933029996,-0.43476204871994967), NE * labelscalefactor); dot((10.882860138479776,-4.139347807671598),linewidth(4.pt) + dotstyle); label("$D$", (10.990208350571054,-3.935767637882502), NE * labelscalefactor); dot((5.076700794792333,-4.714215069422832),linewidth(4.pt) + dotstyle); label("$E$", (5.163359887489334,-4.52334899550419), NE * labelscalefactor); dot((6.554309446328752,-4.567917183132097),linewidth(4.pt) + dotstyle); label("$F$", (6.65679583811112,-4.376453656098768), NE * labelscalefactor); dot((6.741818344632492,1.442337976696709),linewidth(4.pt) + dotstyle); label("$M$", (6.828173734084111,1.6462552595235258), NE * labelscalefactor); dot((5.324966832003627,-0.055651811144906166),linewidth(4.pt) + dotstyle); label("$N$", (5.432668009732607,0.12833675233416722), NE * labelscalefactor); dot((3.0380516090263376,-2.8082418047576243),linewidth(4.pt) + dotstyle); label("$X$", (3.1313076923810037,-2.6137095832337063), NE * labelscalefactor); dot((11.72834945333327,-6.248210432073679),linewidth(4.pt) + dotstyle); label("$Y$", (11.822615273868443,-6.0412675026935485), NE * labelscalefactor); dot((6.717010958622977,-6.211202457303764),linewidth(4.pt) + dotstyle); label("$H$", (6.803691177516541,-6.016784946125978), NE * labelscalefactor); dot((7.21078230814106,-11.198293087436406),linewidth(4.pt) + dotstyle); label("$I$", (7.317824865435516,-11.011226485910319), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $\frac{FD}{FC}= \frac{FP}{FA}= \frac{FE}{FB}$ implies that $EF.FC=BF.FD$. Thus, we get $AP$ is the radical axis of $(ABD)$ and $(ACE)$. The length condition implies$\angle DAX= \angle DAX= \angle DHA$. Let $I$ be the reflection of $A$ over $BC$. So, $\angle FBI= 90- \angle BIA= 90 -\angle BAI= 90- \angle BDH = \angle AHD= \angle ABD= \angle DAX= 90- \angle ADX/2= 90-\angle ABX/2$ Thus, $\angle ABX+ \angle ABD= 180-2\angle FBI+\angle FBI= 180- \angle FBI$. We conclude $I-B-X$ . And by a simialr angle chase we conclude $I-Y-C$. As $AH \cap BX \cap CY= I$, we are done by radical axes theorem that $BCXY$ is cyclic.

[asy][asy] import geometry; import olympiad; defaultpen(fontsize(7pt)); size(250); pair A = dir(105); pair B = dir(210); pair C = dir(330); pair F = foot(A,B,C); pair P = F + 0.34*(A-F); pair D = extension(C-A+P,P,B,C); pair E = extension(B-A+P, P, B,C); pair L = 2*F-A; pair X = intersectionpoints(line(L,B), circumcircle(B,D,A))[0]; pair Y = intersectionpoints(line(L,C), circumcircle(E,C,A))[1]; pair H = intersectionpoints(circumcircle(A,D,B), circumcircle(A,E,C))[1]; draw(A--B--C--A, chartreuse); draw(A--L, blue); draw(X--L,dashed+ blue); draw(Y--L, dashed+blue); draw(A--X, pink); draw(A--Y, magenta); draw(P--D, brown); draw(P--E, brown); draw(D--A, pink); draw(D--X, pink); draw(E--A, magenta); draw(E--Y, magenta); draw(circumcircle(A,B,D), brown); draw(circumcircle(A,E,C), brown); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$F$", F, dir(320)); dot("$P$", P, dir(30)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$L$", L, dir(L)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$H$", H, dir(235)); [/asy][/asy] Very nice problem! Define $BX \cap AF:=L$ Since $PE \parallel AB$ from thales theorem we have $\frac{EF}{FB}=\frac{PF}{AF}$ similarly $PD \parallel AC$ so from thales we have $\frac{FD}{FC}=\frac{FP}{AF} \implies \frac{FD}{FC}=\frac{EF}{FB} \implies FD \times FB= EF \times FC \implies \mathrm{pow}(F,\odot(ABD))=\mathrm{pow}(F,\odot(AEC))=F \in \text{radical axis of} \hspace{0.1cm} \{\odot(ABD), \odot(AEC)\}.$ Since $AXBD$ is cyclic we have $\measuredangle{AXD}=\measuredangle{B}$ and since $DX=AD \implies \measuredangle{ADX}=180^{\circ}-2A =\measuredangle{ABX} \implies \measuredangle{CBX}=\measuredangle{B}$ since $BF \perp AL$ we have $\triangle{ABL}$ is isosceles or in other words , $BC$ is the perpendicular bisector of $AL$. Now since $C$ lies on perpendicular bisector $AL$ and doing similar angle chasing we can show that $\measuredangle{FCL}=\measuredangle{C}$ , hence $BX \cap AP \cap YC=L$. Now we finish by radical center that $B,C,X,Y$ are concylic. $\square$

Let $BX\cap CY=H$. Claim: $AF$ is the radical axis of $(ABD),(ACE)$ Proof: Note that by the parallel condition: \[\angle FEP=\angle FBA,\]\[\angle FDP=\angle FCA,\]meaning that $\triangle PED\sim \triangle ABC$. Therefore, there exists a homothety with center $F$, sending $\triangle PED$ to $\triangle ABC$. Thus: \[\frac{FE}{FB}=\frac{FD}{FC},\]\[FE\cdot FC=FB\cdot FD,\]\[Pow_{(ABD)}(F)=Pow_{(ACE)}(F),\]so $F$ lies on the radical axis $\square$ Claim: $BX,CY,AF$ are concurrent Proof: We work directed angles modulo $180^{\circ}$. We can say: \[\measuredangle ABC=\measuredangle ABD=\measuredangle AXD=\measuredangle XAD=\measuredangle XBD=\measuredangle XBC,\]implying that $XB$ is the reflection of $AB$ about $BC$. Similarly, $CY$ is the reflection of $AC$ about $BC$. As $AF\perp BC$, $HF\perp BC$, so $A,F,H$ are collinear $\square$ Therefore: \[HX\cdot HB=Pow_{(ABD)}(H)=Pow_{(ACE)}(H)=HY\cdot HC,\]implying the result $\blacksquare$

Solved with stqrlight, amoghsus, C9, Jndd, ishan. Claim. $\overline{AF}, \overline{BX}, \overline{CY}$ concur. Proof. Let $P = \overline{XB} \cap \overline{YC}$. Then $\angle PBC = \angle ABD$ by the converse of Fact 5 applied to $(ABX)$, and similarly $\angle ACB = \angle PCB$. It follows that $P$ is the reflection of $A$ over $\overline{BC}$ and said result follows. $\blacksquare$ But $F$ is on the radical axis of $(ABD)$ and $(ACE)$ as $FD \cdot FB = FE \cdot FC$, so the result follows by radical center theorem.

Let $A'$ be the reflection of $A$ across $BC$. Note that $FB \cdot FD = FB \cdot (FC \cdot \frac{FP}{FA})=FC \cdot (FB \cdot \frac{FP}{FA}) = FC \cdot FE$. So $F$ lies the radical axis of $(ABD)$ and $(ACE)$. It follows that $AA'$ is the radical axis of the two circles. Also, $\measuredangle XBD = \measuredangle XAD = \measuredangle DXA = \measuredangle DBA = \measuredangle A'BD$, which implies that $X,B,A'$ are collinear. Similarly, we have that $Y,C,A'$ are collinear. Since $BX$ and $CY$ intersect on the radical axis of $(ABD)$ and $(ACE)$, we have that $B,C,X,Y$ are concyclic by the radical axis theorem.

Is this possible with only angle chasing (no Radical Axis Theorem)?

let the other intersection of the two circles be $Q$ first, power of a point instantly yields that $Q$ lies on $AP$, and therefore $AP$ is a radical axis of the two circles let the extensions of $BX$ and $CY$ be $R$ since $DA=DX$, $DBX=180-DBA$ and $DBA=DBR$ this yields $BA=BR$, and similarly, we can get $CA=CR$ thus, $R$ is a reflection of $A$ over $BC$, and lies on $AP$ therefore, $RX*RB=RC*RY$ by radical axis, and we are done

Note $\triangle ABC\sim\triangle PED$ with $AB/PE=k$. Then, \[EF\cdot FC=\frac{BF}{k}\cdot kFD=BF\cdot FD\]so $F$ lies on the radical axis of $(ABD)$ and $(ACE)$. Letting $Z=\overline{XB}\cap\overline{CY}$, we have $\angle ZBC=180-\angle XBD=\angle ABD$ and similarly $\angle BCZ=\angle C$ so $Z$ is the reflection of $A$ over $F$. Hence, $BXCY$ is cyclic by radical axis. $\square$

dstanz5 wrote: Is this possible with only angle chasing (no Radical Axis Theorem)? Yes

Sketch: Claim: $F$ has equal power with both circles Claim: $A'$, the reflection of $A$ over $BC$, lies on $BX$ and $CY$ Conclude by PoP.

Let $Z = XB \cap YC$. It suffices to prove that $Z$ lies on the radical axis of $(ABD)$ and $(ACE)$. We have \[ \measuredangle CBZ = \measuredangle DAX = \measuredangle AXD = \measuredangle ABC, \]so $BZ$ is the reflection of $BA$ over $BC$. Similarly, $CZ$ is the reflection of $CA$ over $BC$, so $Z$ is the reflection of $A$ over $BC$ and lies on line $AF$. We have \[ BF \cdot FD = BF \cdot FC \cdot \frac{PF}{AF} = EF \cdot FC, \]so line $AF$ is the radical axis of $(ABD)$ and $(ACE)$, and $AF$ passes through $Z$, as desired. $\square$

First of all we claim that $(ABD) \cap (ACE) = T$ lies on $AF$. This is true since from the parallel lines, we have that, $$\frac{FD}{FC} = \frac{FP}{FA} = \frac{FE}{FB}$$which implies that $FE \times FC = FD \times FB$ which implies $F$ lies on the radical axis of $(ABD)$ and $(ACE)$. Suppose $BX \cap CY = Z$, now notice that if $Z$ lies on $AF$, we will be done since then we would have $ZB \times ZX = ZT \times ZA = ZY \times ZC$ which implies the desired result. Now just notice that $\angle ABC = \angle ABD = \angle AXD = \angle XAD = \angle DBZ$ and similarly $\angle ACB = \angle ECZ$, so $Z$ is just the reflection of $A$ across $BC$ and thus lies on $AF$. $\blacksquare$

bored Let $(ABD) \cap (ACE) = Z$ and let $A'$ be the reflection of $A$ over $BC$. Claim: $Z$ lies on $\overline{AF}$ (so that the radical axis of $(ABD)$ and $(ACE)$ is $\overline{AF}$.) Proof. Note that $\triangle ABC$ and $\triangle PED$ are homothetic with center $F$. It follows that \[ \frac{DF}{FE} = \frac{CF}{FB}, \]or $CF \cdot FE = DF \cdot FB$. Thus $F$ has equal power with respect to $(ABD)$ and $(ACE)$, so $\overline{AF}$ is the radical axis of the two circles, as desired. $\square$ Finally, note that \[ \measuredangle XBD = \measuredangle XAD = \measuredangle DXA = \measuredangle DBA = \measuredangle CBA = \measuredangle CBA' \]so $\overline{XBA'}$ collinear. Similarly $\overline{YCA'}$ collinear, hence $A'X \cdot A'B = A'Y \cdot A'C = A'Z \cdot A'A$ or $BCYZ$ cyclic.

No Bash? lol

Easy and a cute one Claim 1: $\color{blue}{AF,BX}$ and $\color{blue}{CY}$ are concurrent. Proof: Note that $\square ABXD$ and $\square ACYE$ are cyclic quadrilaterals. As $DA=DX$ thus, both of these chords subtend equal angles at the center and hence at the circle. Thus, $\angle{ABD}=\angle{XBD}=\angle{XAD}=\alpha$(say) and analogously, $\angle{ACE}=\angle{YCE}=\angle{EAY}=\beta$(say). Also, $\angle{PED}=\alpha$ and $\angle{PDE}=\beta$ since, $PE \parallel AB$ and $PD \parallel AC$. Let $Z$ be the reflection of $A$ over $BC$. Then, $\angle{ZBC}=\alpha$ and $\angle{ZCB}=\beta$ implying, $\angle{ZBD}=\angle{XBD}$ and $\angle{ZCE}=\angle{YCE}$. As $X,Y,Z$ all lie on the same side of $BC$ thus, $AF,BX$ and $CY$ are concurrent at $Z$ as desired.$\blacksquare$ Claim 2: $\color{blue}{AZ}$ is the radical axis of $\color{blue}{(ABXD)}$ and $\color{blue}{(ACYE)}$. Proof: Let $(ABXD) \cap AZ=O$. Then $\angle{AOD}=\angle{ABD}=\alpha$ but we also have $\angle{PED}=\alpha$ thus, $\angle{PED}=\angle{POD}$ implying $P,E,O,D$ lie on a circle. Thus, $\angle{POE}=\angle{PDE}=\beta$ but $\angle{ACE}=\alpha$ thus, $\angle{ACE}=\angle{AOE}$ implying $A,C,Y,O,E$ lie on a circle. Thus we have $(ABXOD)$ and $(ACYOE)$. Clearly $AO$(or $AZ$) is the radical axis of these circles as desired. $\blacksquare$ Claim 3: $\color{blue}{B,C,X,Y}$ are concyclic. Proof: Call $(ABXOD)$ and $(ACYOE)$ as $\Gamma_{1}$ and $\Gamma_{2}$ respectively. Computing powers of $Z$ we get, $$Pow_{(\Gamma_{1})}Z=Pow_{(\Gamma_{2})}{Z} \implies \boxed{ZX.ZB=ZY.ZC}$$implying $B,C,X,Y$ are concylic as desired. $\blacksquare$ ($\mathcal{QED}$) Remark: With trivial angle chasing one can also show that $O$ is the circumcenter of triangle $XYZ$.