Let $G$ be the miquel point of $BDEC$, let $M=(D+B)/2$ and $N=(C+E)/2$, since $GDM \sim GEN$ hence $G$ lies on $(AMN)$ as well similarly if $K=(X+Y)/2$ then $K \in (GAMN)$ and because $AP$ is the diameter we are done.

Construct parallelograms $YDXD'$ and $YEXE'$. Note that $D,E,D',E'$ are trivially collinear. Moreover, let $M,S,T$ be the midpoints of $AX,BD,CE$ respectively. It suffices to prove that $A,M,S,T$ are concyclic, or equivalently that $\angle SMT=\angle A$. Since $BD' \parallel SM$ and $CE' \parallel MT$, we have that $\angle SMT=\angle(BD',CE'),$ and so it suffices to prove that $\angle E'ZD'=\angle A$, where $Z \equiv BD' \cap CE'$. We prove the following Claim: Claim: Triangles $XE'C$ and $XD'B$ are similar. Proof: Note that $\angle E'XC=\angle E'XD'+\angle D'XC=\angle DYE+\angle D'XC=\angle BXC+\angle D'XC=\angle BX'D$. Moreover, $XD'/XE'=YD/YE=BX/XC,$ where the last equality follows since triangles $YDE$ and $XBC$ are similar ($\angle YED=\angle YAD=\angle XCB$ and $\angle YDE=180^\circ-\angle YAX=\angle XBC$) $\blacksquare$ To the problem, since the aforementioned triangles are similar, we obtain that $\angle XD'Z=\angle XE'Z,$ and so $E'D'ZX$ is cyclic, implying that $\angle E'ZD'=\angle D'XE'=\angle DYE=\angle A,$ as desired.

Let $(ADE) \cap (ABC) = K$ Let the midpoint of $BD,CE$ as $M,N$ Let $O$ be the circumcenter of $ABC$ and $O'$ be the circumcenter of $ADE$ Let the circle with the diameter of $AP$ as $\omega$ $\omega \cap AX = Z$ and $AO \cap (ABC) = R$ and $AO' \cap (ADE) = Q$ Let the midpoint of $QR = P'$ It is trivial that the foot of perpendicular line from $P'$ to $AC$ is $N$ Then $P = P'$ and $P,Q,R$ are collinear $\angle AKP = \angle AKR = 90^o$ therefore, $\omega, (ABC), (ADE)$ meets at $A$ and $K$ It is easy to show that $\triangle KDB ~ \triangle KEC ~ \triangle KYX\,(AA)$ Here, $\triangle KDM ~ \triangle KEN ~ \triangle KYZ\,(AA)$ Since $KM$ and $KN$ are the midline of $\triangle KDB,\, \triangle KEC,$ $Z$ is the midpoint of $XY$ Then $\triangle PXZ \equiv \triangle PYZ\,(SAS)$ and $PX = PY$

Rephrase the question as: Circles $C_1,C_2$ intersect at $A,B$. For any point $P$ on $C_1$, consider $P_1=AP\cap C_2$ and $P_2$ the midpoint of $PP_1$. Let $l_P$ be the line through $P_2$ perpendicular to $PP_1$ (the perp bisector of $PP_1$). Show all $l_P$ concur. With this the proof is simple: by spiral from $B$ all the midpoints lie on a circle, and we are done.

simple angle chasing: since $\overline{PB}=\overline{PD}, \overline{PC}=\overline{PE}$, we have $\angle BDP=\angle DBP=\angle ABP=\angle ACP=\angle ECP=\angle CEP$ so $\angle ADP=\pi -\angle BDP=\pi- \angle CEP=\angle AEP$ $\therefore P,A,D,E$ concyclic. then $\angle PXY=\angle ACP=\angle ABP=\angle ADP=\angle BDP=\angle ADP=\angle AYP$ $\therefore \overline{PX} = \overline{PY}$.

gnoka wrote: simple angle chasing: since $\overline{PB}=\overline{PD}, \overline{PC}=\overline{PE}$, we have $\angle BDP=\angle DBP=\angle ABP=\angle ACP=\angle ECP=\angle CEP$ so $\angle ADP=\pi -\angle BDP=\pi- \angle CEP=\angle AEP$ $\therefore P,A,D,E$ concyclic. then $\angle PXY=\angle ACP=\angle ABP=\angle ADP=\angle BDP=\angle ADP=\angle AYP$ $\therefore \overline{PX} = \overline{PY}$. How $\angle{ABP}=\angle{ACP}$ ?

gnoka wrote: simple angle chasing: since $\overline{PB}=\overline{PD}, \overline{PC}=\overline{PE}$, we have $\angle BDP=\angle DBP=\angle ABP=\angle ACP=\angle ECP=\angle CEP$ so $\angle ADP=\pi -\angle BDP=\pi- \angle CEP=\angle AEP$ $\therefore P,A,D,E$ concyclic. then $\angle PXY=\angle ACP=\angle ABP=\angle ADP=\angle BDP=\angle ADP=\angle AYP$ $\therefore \overline{PX} = \overline{PY}$. This is not true or I misunderstood your solution.... $\angle ABP \neq \angle ACP$ maybe you misread the problem....

$\text{solved by using the idea of spiral similarity}$

Let $R,S,T$ be the midpoints of $\overline{BD}$, $\overline{CE}$, $\overline{XY}$ respectively. By the FORGOTTEN COAXIALITY LEMMA (ratio = -1) $ARST$ is cyclic, since $ABXC$ and $ADEY$ are. On the other hand, since $\angle PRA=\angle PSA=90^\circ$ it follows that $\angle PTA=90^\circ$ as well, hence $\triangle PXY$ is $P$-isosceles. $\blacksquare$

pokmui9909 wrote: In a triangle $ABC ~(\overline{AB} < \overline{AC})$, points $D (\neq A, B)$ and $E (\neq A, C)$ lies on side $AB$ and $AC$ respectively. Point $P$ satisfies $\overline{PB}=\overline{PD}, \overline{PC}=\overline{PE}$. $X (\neq A, C)$ is on the arc $AC$ of the circumcircle of triangle $ABC$ not including $B$. Let $Y (\neq A)$ be the intersection of circumcircle of triangle $ADE$ and line $XA$. Prove that $\overline{PX} = \overline{PY}$. Let $ S = (ADE) \cap (ABC) ( S \neq A ) $ U, V, Z in order are midpoint of segments BD, CE, XY Then we have : -- $ (XY, XS) = (XA, XS) = (CA, CS) = (CE, CS) (mod $ $ \pi ) $ , Cuz X, S both lie on (ABC) -- $ (YX, YS) = (YA, YS) = (EA, ES) = (EC, ES) (mod $ $ \pi) $ , Cuz Y, S both lie on (ADE) So, $ \triangle SEC \thicksim^+ \triangle SYX $. Similar : $ \triangle SDB \thicksim^+ \triangle SYX $ Therefore $ \triangle SDU \thicksim^+ \triangle SEV \thicksim^+ \triangle SYZ $ Lead to $ (US, UA) = (VS, VA) = (ZS, ZA) (mod$ $ \pi ) $ ,which means $ A, S, U, V, Z $ are concyclic In additon : Cuz $ PE = PC, PD = PB $ so $ \angle PVA = \angle PUA = 90^o $ Hence, $ \angle PZA = 90^o $ , which means $ PX = PY $

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One-Liner Let $A',A^{*}$ denote the antipodes of $A$ in $(AEF),(ABC)$, then $P$ is mid point of $A'A^{*}$, taking projections to $XY$ we get $PM \perp XY$ where $M$ is mid pt of $XY$. Done

IMO 1979 P3