bgn wrote: $BE=OC$ I guess you mean $PC$?

anantmudgal09 wrote: bgn wrote: $BE=OC$ I guess you mean $PC$? Edited

More general,Given $\triangle ABC$ with circumcircle $\odot O$.$Q$ is an arbitrary point on $\odot O$ .$\odot (B,BQ) \cap AB = \{ E_1,E_2 \}, \odot (C,CQ) \cap AC= \{ F_1,F_2 \}.$ Let $K$ be the projection of $A$ on the Steiner line of $Q$ WRT $\triangle ABC$. Then $K$ is one of theintersection points of $\odot (E_1F_1)$ and $\odot (E_2F_2).$ P.S The solution of this problem is not hard. Let $Q$ be the Euler reflection point ,then we can deduce the problem in post #1.

Suppose that $O'$ be the circumcenter of $BHC$. We know that $O'O=PH,O'H=OP$ so $OHPO'$ is an isosceles trapezoid. Let $D\equiv PM \cap OH$ and $M\equiv OO'\cap BC$. We know that $M$ is midpoint of $OO'$ and $PO' \parallel OD$ so $M$ is midpoint of $PD$ and $BDCP$ is a parallelogram. Now we have $BD=CP=BE,CD=BP=CF \Rightarrow \angle EDF=360-\angle EDB-\angle FDC-\angle BDC=180+\frac{1}{2}\angle EBD+\frac{1}{2}\angle FCD-\angle BDC=180-\frac{1}{2}\angle A-\frac{1}{2}\angle BDC=90$ Suppose that $N$ be the midpoint of $EF$, $Q\equiv \odot ABC\cap \odot AEF$ and $D'$ be the reflection of $D$ wrt $N$. We know that $QB/QC=BE/CF=CP/BP$ so $PQ$ bisects $BC$ or $P,M,D$ and $Q$ are collinear. Now we have $\angle QMN=\angle QCF=\angle QPA$ so $MN\parallel AP$ and $D'$ lies on $AP$. We have $\angle D'ED=90=\angle D'KD$ so $D',E,K,D$ lie on a circle and similarly $D',F,K,D$ lie on a circle and finaly $D',E,F,K,D$ lie on a circle. So $\angle EKF=\angle EDF=90$ and we are done.

Lsway wrote: More general,Given $\triangle ABC$ with circumcircle $\odot O$.$Q$ is an arbitrary point on $\odot O$ .$\odot (B,BQ) \cap AB = \{ E_1,E_2 \}, \odot (C,CQ) \cap AC= \{ F_1,F_2 \}.$ Let $K$ be the projection of $A$ on the Steiner line of $Q$ WRT $\triangle ABC$. Then $K$ is one of theintersection points of $\odot (E_1F_1)$ and $\odot (E_2F_2).$ P.S The solution of this problem is not hard. Let $Q$ be the Euler reflection point ,then we can deduce the problem in post #1. Yes.It's not hard: Let $D$ be the intersection of $\odot (C,CQ) , \odot (B,BQ)$(or the reflection of $Q$ in $BC$) so by simple angle chasing : $\angle E_1DF_1=\angle E_2DF_2=90$ so $D$ be the second intersection of $\odot (E_1F_1)$ and $\odot (E_2F_2).$(or the intersection of $E_1F_2,E_2F_1$) Let $M,N$ be the midpoints of $E_1F_1,E_2F_2$. so we need to prove that $MN \parallel AK$ which is well-known and obvious (because by simple angle chasing ($\angle DCB=\angle BCQ=\angle CBQ'$) $DCQ'B$ be the parallelogram $.$ ($Q'$ be the reflection of $Q$ in perpendicular bisector of $BC$) let $M',M''$ be the midpoints of $DQ',KD$ so $Q'K \parallel M'M'' \parallel AK$ and also $M,N,M',M''$ are conllinear so done$.$)$.$ so we are done.$\blacksquare$.

I notice the similarity of a part of statement of the topic https://artofproblemsolving.com/community/c6t48f6_geometry. i will use the same idea : Let $G$ the point of intersection of $(AEF)$ and of $(ABC)$ .$P'$ the point of intersection of $AP$ and $(AEF)$. $I,J$ the midpoints of $BC,EF$, we know that $G$ is the similicenter of the similarity that sends $B,C,I,P$ to resp. $E,F,J,P'$ and $\frac{P'F}{P'E}=\frac{PC}{PB}=\frac{BE}{CF}=\frac{GE}{GF} \implies GP' $ pass through $J \iff$ $G,J,P'$ are collinear then $G,I,P$ as well. hence $IJ \parallel P'P$ .In the other hand consider $L$,the reflection of $P$ in $I$ it's easy to show that $\angle ELF=90^{\circ} \wedge L\in OH$ .we have $IJ \parallel KP $ thus $IJ\perp KL\equiv OH $ then $IJ$ bisects $KL$ so $J$ is equidistant from $K,L$ i.e. $EFKL$ is cyclic therefore $\angle EKF = 90 ^{\circ}$. RH HAS

I took more than an hour. Mostly because the Euler-line itself is a red-herring. Not that it's not useful, but it serves mainly as a distraction; dealing with $E, F$ is the actual difficulty of the problem. That said, it's a nice problem! Iran TST 2017 wrote: In triangle $ABC$ let $O$ and $H$ be the circumcenter and the orthocenter. The point $P$ is the reflection of $A$ with respect to $OH$. Assume that $P$ is not on the same side of $BC$ as $A$. Points $E,F$ lie on $AB,AC$ respectively such that $BE=PC \ , CF=PB$. Let $K$ be the intersection point of $AP,OH$. Prove that $\angle EKF = 90 ^{\circ}$ Proposed by Iman Maghsoudi

Construct parellogram BPCB', (AEF) cuts (O)=G. SS: GB/GC=BE/CF=PB/PC -> G,D,P collinear. SS send J to D -> JD//IP hence JD⊥OH. P'KP=90 so DP'=DK hence JP'=JK. Angle chase: EP'F=90, BP'E and CP'F is isoceles -> JF=JE=JP'=JK => (EP'KF) is cyclic so EKF=90 (qed)

Excellent problem. [asy][asy] size(7.5cm); defaultpen(fontsize(10pt)); pen pri=heavyblue; pen sec=purple; pen tri=red; pen qua=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pair O,A,B,C,H,P,D,EE,F,K,M,Y,NN; O=(0,0); A=dir(115); B=dir(215); C=dir(325); H=A+B+C; P=reflect(O,H)*A; D=B+C-P; EE=intersectionpoint(A--B,circle(B,length(C-P))); F=intersectionpoint(A--C,circle(C,length(B-P))); K=(A+P)/2; M=(B+C)/2; Y=2*foot(O,P,D)-P; NN=(EE+F)/2; filldraw(circumcircle(A,EE,F),qfil,qua); draw(Y--P,qua); filldraw(circumcircle(K,EE,F),tfil,tri); draw(EE--F,tri); filldraw(circumcircle(D,K,P),sfil,sec); filldraw(B--P--C--D--cycle,sfil,sec); draw(M--NN,sec+dashed); draw(A--P,sec); filldraw(circumcircle(B,H,C),fil,pri); filldraw(circle(O,1),fil,pri); draw( (5H-4O)--(5O-4H),pri); real t=1.28; clip( (-t,-t)--(-t,100)--(t,100)--(t,-t)--cycle); draw(A--B--C--A,pri); dot("$O$",O,dir(250)); dot("$A$",A,N); dot("$B$",B,W); dot("$C$",C,E); dot("$H$",H,S); dot("$P$",P,SE); dot("$D$",D,dir(285)); dot("$E$",EE,dir(120)); dot("$F$",F,dir(30)); dot("$K$",K,dir(75)); dot("$M$",M,S); dot("$Y$",Y,SW); dot("$N$",NN,NW); [/asy][/asy] Let $M$ be the midpoint of $\overline{BC}$, $N$ the midpoint o $\overline{EF}$, $D$ the point such that $BPCD$ is a parallelogram, and $Y$ the second intersection of $\overline{PMD}$ with $(ABC)$. Claim 1. $D$ lies on $\overline{OH}$. Proof. By reflection through $M$, $D$ lies on $(BHC)$. Thus \[\measuredangle BHD=\measuredangle BCD=\measuredangle CBP=\measuredangle CAP=90^\circ+\measuredangle(\overline{AC}+\overline{OH})=\measuredangle BHO,\]as desired. $\blacksquare$ Claim 2. $\angle EDF=90^\circ$. Proof. Since $BE=PC=BD$ and $CF=PB=CD$, \begin{align*} \measuredangle EDF&=\measuredangle EDB+\measuredangle BDC+\measuredangle CDF=\measuredangle BED+\measuredangle CPB+\measuredangle DFC\\ &=\measuredangle AED+\measuredangle FAE+\measuredangle DFA=\measuredangle FDE. \end{align*}Since $D\notin\overline{EF}$, it follows that $\angle EDF=90^\circ$. $\blacksquare$ Claim 3. $Y$ is the Miquel point of $BCFE$; in particular, $Y\in(AEF)$. Proof. Since $M\in\overline{YP}$, we have \[\frac{YB}{YC}=\frac{PC}{PB}=\frac{BE}{CF},\]whence $\triangle YBE\sim\triangle YCF$, as desired. $\blacksquare$ Claim 4. $\overline{MN}\perp\overline{OH}$. Proof. A spiral similarity at $Y$ sends $\overline{BC}\cup M$ to $\overline{EF}\cup N$, so \[\measuredangle YMN=\measuredangle YBE=\measuredangle YBA=\measuredangle YPA=\measuredangle(\overline{YM},\overline{AP}),\]it follows that $\overline{MN}\parallel\overline{AP}$, which is sufficient. $\blacksquare$ Finally since $\angle DKP=90^\circ$, $M$ is the center of $(DKP)$, whence $D$ and $K$ are reflections across $\overline{MN}$. It follows that $K\in(DEF)$, so $\angle BKC=\angle BDC=90^\circ$, and we are done.

Excellent problem. Let $D$ be the point such that $BDCP$ is a parallelogram. Let $M$ denote the midpoint of $BC$. Claim : $D \in OH$.

Claim : $\angle EDF=90^{\circ}$. Proof : We have : $$ \angle EDF = 2\pi - \{\angle EDB + \angle BDC + \angle FDC \}$$$$\iff \angle EDF = 2\pi - \left(\frac {\pi}2 - \frac {\angle B - \angle PCB}2 \right) + \left ( \pi-\angle A \right) + \left(\frac {\pi}2 - \frac {\angle C - \angle PBC}2 \right) $$$$ \angle EDF = \frac {\angle B}2 + \frac {\angle C}2 +\angle A - \left( \frac {\angle PBC +\angle PCB}2 \right) = \frac {\angle B}2 + \frac {\angle C}2 +\frac {\angle A}2 = \frac {\pi}2$$ Let $N$ denote the midpoint of $EF$. Claim : $N \in \odot (BC)$. Proof : We have : $$\frac {\pi}2 = \angle (ED,DF) = \angle (\perp ED , \perp DF) = \angle (BN,NC)$$ Claim : $MN \parallel AP$. Proof : Suppose $AP \cap BC = G$ and $\angle PBC= \alpha$. We need to prove that $\angle AGB = \angle NMB$. We have : $$ \angle NMB = 2\angle NCB = 2(\angle DCB + \angle NCF) = 2\left( \alpha + \frac {\angle C - \alpha}2\right)= \alpha + \angle C = \alpha + \angle APB = \angle AGB$$ So the claim holds. $\blacksquare$ At this point , note that $DK\perp KP \implies K\in \odot (PK)$. Hence , $M$ is the circumcenter of $\odot (KDP)$. Note that since $MN$ is perpendicular to $DK$ and $MD=MK$ , this means that $MN$ is the perpendicular bisector of $DK$. This in turn means $ND=NK=NE=NF \implies K\in \odot (EF)$. We are done. $\blacksquare$

Basically the same solution as all of the above. Let $M$ be the midpoint of segment $BC$ and let $P'$ be the reflection of $P$ over $M$. Let $A'$ be the $A$-antipode. Then since $HPA'P'$ is a parallelogram, $HP'\parallel PA'$. But $OH \parallel PA'$, so it follows that $P'$ lies on $\overline{OH}$. Let the internal angle bisectors of $\angle ABP'$ and $\angle ACP'$ meet at $D$. Then \[\angle DBC+\angle DCB=\frac{\angle B+\angle P'BC}{2}+\frac{\angle C+\angle P'CB}{2}=\frac{\angle A+\angle B+\angle C}{2}=90^\circ,\]so $\angle BDC=90^\circ$. This implies that $\angle EP'F=90^\circ$ as well, and since $D$ is its circumcenter, it follows that $D$ is the midpoint of $EF$. Now angle chasing shows that \[\angle(BD,AP)=\measuredangle BAP+\measuredangle DBA=\measuredangle CBP'+\measuredangle P'BD=\measuredangle CBD=\measuredangle BDM,\]so $DM\perp OH$. Since $M$ lies on the perpendicular bisector of $P'K$, it follows that $D$ also lies on this line, so $DP'=DK$. Consequently $K$ lies on the circle with diameter $EF$ and hence $\angle EKF=90^\circ$. [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(130); pair B = dir(215); pair C = dir(325); pair H = orthocenter(A,B,C); pair O = circumcenter(A,B,C); pair K = foot(A,O,H); pair P = 2*K-A; pair M = midpoint(B--C); pair P1 = B+C-P; pair E[] = intersectionpoints(Circle(B,abs(P1-B)),L(A,B)); pair F[] = intersectionpoints(Circle(C,abs(P1-C)),L(A,C)); pair D = midpoint(E[0]--F[0]); draw(E[0]--A--F[0], black+1); draw(B--C, black+1); draw(A--P); draw(P--C, fuchsia+1); draw(P--B, royalblue+1); draw(B--E[0], fuchsia+1); draw(C--F[0], royalblue+1); draw(P--P1, mydash); draw(B--P1, fuchsia+1); draw(C--P1, royalblue+1); draw(O--H); draw(B--D--C, mydash); draw(E[0]--F[0]); draw(D--M); draw(circumcircle(A,B,C)); draw(circumcircle(E[0],F[0],K), dotted); draw(rightanglemark(A,K,P1,1.5)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$H$", H, dir(180)); dot("$O$", O, dir(30)); dot("$P'$", P1, dir(90)); dot("$E$", E[0], dir(180)); dot("$F$", F[0], dir(30)); dot("$D$", D, dir(90)); dot("$K$", K, dir(250)); dot("$M$", M, dir(90)); dot("$P$", P, dir(P)); [/asy][/asy]

Borrowing diagram of @above. Let $P'$ be reflection of $P$ in midpoint $M$ of $BC$. And $D$ be midpoint of $EF$. Let $\Gamma$ denote the composition of homotheties $M(-1) \circ A(2)$, then $\Gamma$ is itself a homothety centered at $G$. As $\Gamma(K) = P'$ we have $P' \in \overleftrightarrow{KG} = \overleftrightarrow {OH}$ . Note that $BE = PC = P'B$ and similarly $CF = CP'$, hence $$\angle EP'F = 360 - \angle BP'E - \angle BP'C - \angle CP'F = 360 - (90 - \frac{\angle P'BA}{2})-\angle BP'C - (90 - \frac{\angle P'CA}{2}) = 180 - \angle BP'C +\frac{\angle P'BA+\angle P'CA}{2} $$$$= 180 - \angle BP'C +\frac{\angle BP'C- \angle BAC}{2} = 180- \frac{\angle BAC+\angle BP'C}{2}=180- \frac{\angle BAC+\angle BPC}{2} = 90 $$So $\angle EP'F = 90$, hence it is enough to show that $K \in (EP'F)$, but this is equivalent to showing that $D$ lies on perpendicular bisector of $P'K$. As $M$ lies on perpendicular bisector of $P'K$ it is enough to show $DM \perp OH$. We do this using vectors . $\overrightarrow v = 2 \overrightarrow {MD} = \overrightarrow {MB} + \overrightarrow {BE} + \overrightarrow {MC} + \overrightarrow {CF} = \overrightarrow{BE}+\overrightarrow{CF} \implies \overrightarrow v\cdot \hat {BC} = BE cos B - CF cos C = CP cos B - PB cos C$ As $\angle BPA = C, \angle CPA = B$ we get $\overrightarrow v\cdot \hat {BC} = \overrightarrow{BC} \cdot \hat {PA}$ And similarly $\overrightarrow v \cdot \hat {BC}_{\perp} = \overrightarrow{BC} \cdot \hat{PA}_{\perp} \implies \angle (v, BC) =tan^{-1} \frac{\overrightarrow v \cdot \hat {BC}_{\perp}}{\overrightarrow v\cdot \hat {BC}}= tan^{-1}\frac{\overrightarrow{BC} \cdot \hat{PA}_{\perp}}{ \overrightarrow{BC} \cdot \hat {PA}} = \angle(BC,PA)$ Hence $PA \parallel v \implies v \perp OH$ as required . Hence proved .

This is really cool. I'll just post an attachment with many of the same ideas as the other solutions, except that I used centroid ratios to prove that $O$, $H$, and $P'$ are collinear, and I finished with projecting $E$ and $F$ onto $\overline{BC}$ and using the law of sines.

After finding this solution, I am not sure whether my geo intuition is either very good or very bad. Lots of details omitted so that it looks like my solution is short. Let $P'$ be the reflection of $P$ over the perpendicular bisector of $BC$. Then $BE=BP'$ and $CF=CP'$. Angle chasing shows that $\measuredangle FP'E$ is thus $90-\measuredangle BAC$. We prove more generally that for any $E$ on $\overline{AB}$ and $F$ on $\overline{AC}$ such that $\measuredangle FP'E=90-\measuredangle BAC$, $\measuredangle FKE=90$. This is a degree $2$ statement by moving points, so we need $3$ points $E$ that work. Let $E_1=\overline{HO}\cap\overline{AB}$. Then we want to show that $\measuredangle AP'E_1=90-\measuredangle BAC$. This is true because it can be shown that $\overline{E_1P'}$ is the reflection of $\overline{HO}$ over $\overline{AB}$, and then angle chasing finishes. Similarly, let $F_2=\overline{HO}\cap\overline{AC}$, then we want to show that $\measuredangle F_2P'A=90-\measuredangle BAC$, which is identical to the first case.

Thus if $F_3=\overline{PK}\cap\overline{AC}$, we have that $\measuredangle F_3KE_3=90, \measuredangle F_3P'E_3=90-\measuredangle BAC$, as desired.

Lemma :Given a triangle $\triangle ABC$ with its circumcircle $\omega$. Suppose that $D$ is an arbitrary point on side $BC$ and $AD$ intersects $\omega$ at point $P$. Let $E$ , $F$ be points on sides $AB$ and $AC$ respectively such that $BE=PC$ and $CF=PB$ , then if bisectors of angles $\angle BDP$ and $\angle CDP$ intersects $AB$ , $AC$ at points $X$ , $Y$ , we have $EF \parallel XY$. Proof : Firstly , since point $X$ is the intersection point of external bisector of angle $\angle ADB$ and line $AB$ , then one can see that : $$\frac{XB}{XA}=\frac{BD}{AD} \implies AX=\frac{AB.AD}{AD-BD} , AY=\frac{AC.AD}{AD-CD} \implies P \iff \frac{AX}{AY}=\frac{AE}{AF} \iff \frac{AB}{(AD-BD)(AB-CP)}=\frac{AC}{(AD-CD)(AC-BP)} $$$$\frac{AB}{AC}.\frac{PB}{PC}=\frac{BD}{CD} \implies AB.PB.CD=AC.CP.BD $$$$\triangle ABD \sim \triangle CDP \implies AB.CD=AD.CP , \triangle ACD \sim \triangle BDP \implies AC.BD=AD.BP$$Thus , the lemma is proved. Now suppose that the circumcircle of triangle $\triangle BHC$ intersects the Euler line of triangle $\triangle ABC$ at point $Q$ for second time , then if $H_B$ , $H_C$ are feet of altitudes from $B$ and $C$ to the sides , while quadrilateral $AH_BKH$ is cyclic , one can see that : $$\angle QCB=\angle QHB=\angle H_BHK=\angle H_BAK=\angle CAP=\angle CBP \implies QC \parallel BP , QB \parallel CP$$Thus the quadrilateral $BQCP$ is a parallelogram. So if liens $BQ$ and $CQ$ intersect circle $\omega$ at points $X$ and $Y$ for second time respectively , then quadrilaterals $CPBX$ and $BPCY$ are isosceles trapezoids and as the result , points $B$ , $C$ are circumcenters of triangles $\triangle EYQ$ and $\triangle FXQ$ respectively and one can see that : $$\angle FQX=\frac{\angle FCX}{2}=\frac{\angle B-\angle BAP}{2} , \angle EQY=\frac{\angle C- \angle CAP}{2} \implies \angle FQX + \angle EQY=90 - \angle A \implies \angle EQF=90 (I)$$Now suppose that the external bisectors of angles $\angle CDP$ , $\angle BDP$ intersects sides $AC$ , $AB$ at points $X'$ , $Y'$ and also lines parallel to $DY'$ , $DX'$ and passes trough points $E$ , $F$ intersect each other at point $T$ , so by lemma we have $X'Y' \parallel EF$ and by homothety of triangles $\triangle ETF$ and $\triangle Y'DX'$ , point $T$ lies on the line $AP$. So if the Euler line intersects the line $BC$ at point $S$ , since quadrilateral $AKH_AS$ is cyclic we have : $$\angle QSB=\angle KAH , \angle QBC=\angle BAP \implies \angle XQK=\angle SQB=90- \angle B , \angle FQX=\frac{\angle B - \angle BAP}{2}$$$$\implies \angle FQK=90-\frac{\angle B + \angle BAP}{2}=180 - \angle ADX'=\angle FTK$$Thus quadrilateral $FTQK$ and similarly $ETQK$ are cyclic and by (I) , we have $\angle FKE=\angle FQE=90$ and we're done.

Cute! Denote by $D$ the midpoint of $EF$, let $P'$ be the point s.t. $P'BCP$ is parallelogram, let $A'$ be the antipode of $A$ in $(ABC)$, Notice we have $P' \in \overline{OH}$ because $P'H \parallel A'H \parallel OH$. Next we notice that the angle bisectors of $EBP',FCP'$ concur at $D$ and $\measuredangle EP'F = 90^{\circ}$ so to finish just notice that $MD$ is the perpendicular bisector of $P'K$.