First a trivial Lemma: consider two arbitrary lines $\ell_1,\ell_2$ and consider three arbitrary points $A,B,C$ then let $A_1,A_2$ be the projections of $A$ on $\ell_1,\ell_2$ respectively and let $A'$ be the midpoint of it. we define $B',C'$ similarly then $A',B',C'$ are collinear if and if only $A,B,C$ are collinear. Proof: it's so easy just observe that using vectors and thales theorem we have $\overrightarrow{A'B'}=\frac{1}{2}(\overrightarrow{A_1B_1}+\overrightarrow{A_2B_2}),\dots\Longleftrightarrow \frac{A_1B_1}{A_2B_2}=\frac{B_1C_1}{B_2C_2}$. Back to original problem: Let the perpendiculars from $E,F$ to $AC,AB$ intersect $AB,AC$ at $B',C'$ respectively and $H$ be the orthocenter of $AB'C'$, $Q,M$ the midpoint of $EF,BC$ and $T,S$ be the antipode of $A$ WRT $\odot(AB'C'),\odot(ABC)$ respectively. note that $K$ is midpoint of $B'C'$ and $K,T,H$ are collinear (well known). Let $TK$ cut $\odot(AB'C')$ again at $P'$ since $\angle TP'A=90$ we get that $P'\in \odot(AH)$ now from the lemmma for two lines $AB,AC$ and three collinear points $Q,M,K$ we deduce that $S\in TP$ hence $P'\in \odot(ABC)\Longrightarrow P=P'$. Q.E.D

hem ,Q,M,K colinear not Q,M,T and this cost my time one hour

It's fairly straight forward after this (For starters, see that sine rule will work nicely after this)

what is antipode, can someone tell me?

Reynan wrote: what is antipode, can someone tell me? The antipode of a point is the point which is diametrically opposite to it.

Solution for iranians;

madmathlover wrote: Reynan wrote: what is antipode, can someone tell me? The antipode of a point is the point which is diametrically opposite to it. can you tell me more clearly? what is diametrically opposite? sorry i am noob:(

Reynan wrote: madmathlover wrote: Reynan wrote: what is antipode, can someone tell me? The antipode of a point is the point which is diametrically opposite to it. can you tell me more clearly? what is diametrically opposite? sorry i am noob:( Let $\Omega$ be a circle.$AA_{1}$ is a diameter of the circle.Then $A_{1}$ is diametrically opposite to $A$ and vice versa.

Any different solution?

Let $U,V$ denote midpoints of segments $A'B,A'C$ where $A'$ is antipode of $A$ WRT $\odot(AEF).$ Since $\Delta EUM$ and $\Delta MVF$ are congruent, it follows that $\angle A'BE=\angle A'CF$ and from here it follows that $\Delta EBA' \sim \Delta A'CF.$ But recall that $P$ is center of spiral similarity mapping $BE \mapsto FC$ hence $\Delta PEB \sim \Delta PFC.$ On account of these 2 similarities it follows that $$\frac{PE}{PF}=\frac{EB}{FC}=\frac{EA'}{A'F}$$Hence quadrilateral $PEA'F$ is harmonic, so it follows that $P,A',K$ are collinear, as desired.

Problem. Let $ABC$ be a triangle inscribed in circle $(O)$ with median $AM$. $E,F$ lie on $CA,AB$ such that $ME=NF$. The tangents at $E,F$ of $(AEF)$ interesect at $S$. $(AEF)$ cuts $(O)$ again at $G$. Prove that $\angle AGS=90^\circ$. Proof. Let $N$ be the midpoint of $EF$ then $M,N,S$ are collinear. The tangents at $B,C$ of $(O)$ intersect at $T$. The spiral similarity center $G$ transform $N,E,F,S$ to $M,B,C,T$, reps. We get $\angle GST=\angle GNM=\angle GMT$ so $GSMT$ is cyclic. Combine with property of symmedian we get angles chasing $$\angle BGT=\angle MGC=\angle BMG-\angle MCG=\angle BMN-\angle GMN-\angle MCG=\angle BMN-\angle GCE-\angle MCG=\angle TMN-90^\circ-\angle BCA=180^\circ-\angle SGT-90^\circ-(180^\circ-\angle BGA)=\angle BGT+\angle SGA-90^\circ.$$We deduce $\angle AGS=90^\circ$.

General problem. Let $ABC$ be a triangle inscribed in circle $(O)$. $M$ is a point on perpendicular bisector of $BC$ and inside triangle $ABC$. $E,F$ lie on $CA,AB$ such that $ME=MF$. $N$ lie inside triangle $AEF$ such that $\triangle NEF\sim\triangle MBC$. $(K)$ is circumcircle of triangle $AEF$. The tangents at $E,F$ of circles $(KNE),(KNF)$ interesect at $S$. $(K)$ cuts $(O)$ again at $G$. Prove that $\angle AGS=90^\circ$.

Let $M$ be the midpoint of $EF,O$ be the center of $(PAFE),$ and $Q\neq P$ be the intersection of $PK$ with $(PAFE).$ Lemma 1: If $A,B,C,D$ are points such that $AC\cap BD=X,$ then $(ACX)\cap (ABX)=P\neq X$ is the spiral center sending $AB$ to $CD.$ Proof: Well known. $\Box$ By Lemma 1, the spiral similarity $\phi$ sending $EF$ to $BC$ is centered at $P,$ so let $\phi(Q)=Q'$ and $\phi(M)=M'.$ Since $PQ$ is the $P$-symmedian chord in $\triangle PEF,$ it's well known that $POMQ$ is cyclic; however, $O$ also lies on $MM',$ so $O,Q,Q'$ have to be collinear again due to Lemma 1. Now it suffices to show that $A,Q,Q'$ are collinear, but this is trivial: $$\measuredangle EAQ=\measuredangle EFQ=\measuredangle BCQ'=\measuredangle BAQ'=\measuredangle EAQ'$$from $\phi(\triangle EFQ)=\triangle BCQ',$ so we're done. $\blacksquare$ [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(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.58, xmax = 7.58, ymin = -5.77, ymax = 5.77; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-0.78,5.29)--(-3.42,-3.69), linewidth(1) + wrwrwr); draw((-3.42,-3.69)--(4.82,-3.69), linewidth(1) + wrwrwr); draw((4.82,-3.69)--(-0.78,5.29), linewidth(1) + wrwrwr); draw(circle((0.7,-0.02316258351893081), 5.515441654020607), linewidth(1) + wrwrwr); draw(circle((-0.6184291103248956,2.2417049276249013), 3.0525739959346954), linewidth(1) + wrwrwr); draw((0.37780385192440746,-2.2404169589134835)--(-3.1893652954799423,3.8874555565225872), linewidth(1) + wrwrwr); draw((-3.1893652954799423,3.8874555565225872)--(-0.78,5.29), linewidth(1) + wrwrwr); draw((-2.4040259437765394,-0.23414885420959397)--(2.047838573787419,0.7553588584623171), linewidth(1) + wrwrwr); draw((2.047838573787419,0.7553588584623171)--(0.7,-3.69), linewidth(1) + wrwrwr); draw((0.7,-3.69)--(-2.4040259437765394,-0.23414885420959397), linewidth(1) + wrwrwr); draw((-0.78,5.29)--(-0.21005095760194714,-5.463006689141201), linewidth(1) + wrwrwr); draw((-0.6184291103248956,2.2417049276249013)--(0.7,-3.69), linewidth(1) + wrwrwr); draw((-0.17809368499456024,0.26060500212636156)--(-0.45685822064979176,-0.8065901447501977), linewidth(1) + wrwrwr); draw((0.7,-3.69)--(-0.21005095760194714,-5.463006689141201), linewidth(1) + wrwrwr); draw(circle((-3.5070639537967256,0.5601671564337461), 3.342421358066939), linewidth(1) + wrwrwr); draw((-0.6184291103248956,2.2417049276249013)--(-2.4040259437765394,-0.23414885420959397), linewidth(1) + wrwrwr); draw((-0.6184291103248956,2.2417049276249013)--(2.047838573787419,0.7553588584623171), linewidth(1) + wrwrwr); draw((-3.1893652954799423,3.8874555565225872)--(-0.21005095760194714,-5.463006689141201), linewidth(1) + wrwrwr); draw((2.047838573787419,0.7553588584623171)--(0.37780385192440746,-2.2404169589134835), linewidth(1) + wrwrwr); draw((0.37780385192440746,-2.2404169589134835)--(-2.4040259437765394,-0.23414885420959397), linewidth(1) + wrwrwr); /* dots and labels */ dot((4.82,-3.69),dotstyle); label("$C$", (4.9,-3.49), NE * labelscalefactor); dot((-3.42,-3.69),dotstyle); label("$B$", (-3.34,-3.49), NE * labelscalefactor); dot((0.7,-3.69),linewidth(4pt) + dotstyle); label("$M'$", (0.78,-3.53), NE * labelscalefactor); dot((-0.78,5.29),dotstyle); label("$A$", (-0.7,5.45), NE * labelscalefactor); dot((-2.4040259437765394,-0.23414885420959397),dotstyle); label("$E$", (-2.32,-0.03), NE * labelscalefactor); dot((2.047838573787419,0.7553588584623171),linewidth(4pt) + dotstyle); label("$F$", (2.12,0.91), NE * labelscalefactor); dot((-3.1893652954799423,3.8874555565225872),linewidth(4pt) + dotstyle); label("$P$", (-3.1,4.05), NE * labelscalefactor); dot((0.37780385192440746,-2.2404169589134835),linewidth(4pt) + dotstyle); label("$K$", (0.46,-2.09), NE * labelscalefactor); dot((-0.21005095760194714,-5.463006689141201),linewidth(4pt) + dotstyle); label("$Q'$", (-0.14,-5.31), NE * labelscalefactor); dot((-0.6184291103248956,2.2417049276249013),linewidth(4pt) + dotstyle); label("$O$", (-0.54,2.41), NE * labelscalefactor); dot((-0.45685822064979176,-0.8065901447501977),linewidth(4pt) + dotstyle); label("$Q$", (-0.38,-0.65), NE * labelscalefactor); dot((-0.17809368499456024,0.26060500212636156),linewidth(4pt) + dotstyle); label("$M$", (-0.1,0.43), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

If $ AT $ is diameter of circle $AEF$ we claim that $\angle {TBA} =\angle {TCA} $ . Proof of the claim : if it's not happen ; get $L , L'$ such that $\angle{TBL} = \angle{TCA} , \angle{TCL'} = \angle{TBA} , \angle{TLB} = \angle{TL'C} = 90 $ It's easy to check that $ME = MF = ML = ML'$ and $ \triangle {TEL} \sim \triangle{TFL'} $ so we can check that $ \triangle {ETF} \cong \triangle {L'TL} $ so $BT=CT$ and it get us $\angle{B} =\angle{C}$ , and the claim is proved . know from the claim we know $\frac{TE}{TF} =\frac {BE}{CF} = \frac {PE}{PF} $ so $(EF,PT)=-1$ , so $P , T , K $ are collinear $\blacksquare$

Arefe wrote: If $ AT $ is diameter of circle $AEF$ we claim that $\angle {TBA} =\angle {TCA} $ . Proof of the claim : if it's not happen ; get $L , L'$ such that $\angle{TBL} = \angle{TCA} , \angle{TCL'} = \angle{TBA} , \angle{TLB} = \angle{TL'C} = 90 $ It's easy to check that $ME = MF = ML = ML'$ and $ \triangle {TEL} \sim \triangle{TFL'} $ so we can check that $ \triangle {ETF} \cong \triangle {L'TL} $ so $BT=CT$ and it get us $\angle{B} =\angle{C}$ , and the claim is proved . know from the claim we know $\frac{TE}{TF} =\frac {BE}{CF} = \frac {PE}{PF} $ so $(EF,PT)=-1$ , so $P , T , K $ are collinear $\blacksquare$ i dont find it's easy to check that $ME=MF=ML=ML'$, hope that someone can make it easy to understand.

xst wrote: Arefe wrote: If $ AT $ is diameter of circle $AEF$ we claim that $\angle {TBA} =\angle {TCA} $ . Proof of the claim : if it's not happen ; get $L , L'$ such that $\angle{TBL} = \angle{TCA} , \angle{TCL'} = \angle{TBA} , \angle{TLB} = \angle{TL'C} = 90 $ It's easy to check that $ME = MF = ML = ML'$ and $ \triangle {TEL} \sim \triangle{TFL'} $ so we can check that $ \triangle {ETF} \cong \triangle {L'TL} $ so $BT=CT$ and it get us $\angle{B} =\angle{C}$ , and the claim is proved . know from the claim we know $\frac{TE}{TF} =\frac {BE}{CF} = \frac {PE}{PF} $ so $(EF,PT)=-1$ , so $P , T , K $ are collinear $\blacksquare$ i dont find it's easy to check that $ME=MF=ML=ML'$, hope that someone can make it easy to understand. oh, i understand, it's a litte lemma. but you don't need to make $L$ and $L'$, when you line $TB$, $TC$ and make their midpoints X, Y you can easily find that $\triangle MXE \cong \triangle MYF$, then we get $\angle XBE = \angle YCF$.

K.N wrote: Let $ABC$ be an arbitrary triangle. Let $E,E$ be two points on $AB,AC$ respectively such that their distance to the midpoint of $BC$ is equal. Let $P$ be the second intersection of the triangles $ABC,AEF$ circumcircles . The tangents from $E,F$ to the circumcircle of $AEF$ intersect each other at $K$. Prove that : $\angle KPA = 90$ Here let take $F \in AB$ and $ E \in AC$ Let $M$ and $N$ midpoints of $BC$ and $EF$ respectively. Let $l$ the angle bisector of $\angle FPC$ (which is also the angle bissector of $\angle BPE$) Let $\phi$ the inversion centered at $P$ radius $\sqrt{PF.PC}$ followed by a reflection through $l$. Since there's a spiral similarity on P who takes $\triangle PEF$ on $\triangle PBC$ then $\phi (F)= C$ and $\phi (E)= B$. Let $M'$ the second intersection of $PK$ with $(PEF)$. So $\phi (M)= M'$. Let $N' \in (PBC)$ such that $PBN'C$ is harmonic quadrilateral. Then $\phi (N)=N'$. Since theres a spiral similarity centered at $P$ who takes $\triangle PFB$ on $\triangle PM'N'$, then $\angle PBF=\angle PN'M'$ which give us that $A, M', N'$ collinear. Let $A'$ the intersection of $EF$ and $BC$ ($\phi (A)=A'$). So we have $PA'MN$ cyclic, then $\angle MPA'= \angle MNA'= \frac{\pi}{2}$ so $\angle M'PA = \frac{\pi}{2}$ .

I'll post my solution, which is similar to the solutions above. Let $M$ and $N$ be the midpoints of $BC$ and $EF$ respectively. Since $PK$ is the $P-$symmedian in $\triangle PEF$ and $P$ is the center of spiral similarity that maps $(E,N,F)\mapsto(B,M,C)$, $$\measuredangle KPF=\measuredangle EPN=\measuredangle(PE,PN)=\measuredangle(BE,MN).$$Since $APEF$ is cyclic, $$\measuredangle FPA=\measuredangle FEA=\measuredangle(EF,EA)=\measuredangle(EF,BE).$$Therefore, $$\measuredangle KPA=\measuredangle KPF+\measuredangle FPA=\measuredangle(BE,MN)+\measuredangle(EF,BE)=\measuredangle(EF,MN)=90^\circ.$$ [asy][asy] size(8cm); pointpen=black; pair A = dir(120), B = dir(210), C = dir(330); pair M = midpoint(B--C); pair[] D = IPs(CR(M, 1), B--A--C); pair E = D[1], F = D[0], N = midpoint(E--F); pair P = (C*E-B*F)/(C+E-B-F); path w = circumcircle(A, E, F); pair O = circumcenter(A, E, F); pair K = extension(E, (0,1)*(O-E)+E, F, (0,1)*(O-F)+F); D(A--B--C--cycle, red); DPA(w^^unitcircle, paleblue); D(E--K--F, orange); markscalefactor=.008; DPA(E--F^^M--N^^rightanglemark(M, N, E), lightblue); D("A", A, A); D("B", B, B); D("C", C, C); D("M", M, S); D("E", E, NE); D("F", F, NE); D("N", N, NE); D("P", P, P); D("K", K, S); [/asy][/asy]

Let $N$ be midpoint of $EF$ and $NK$ meet $AB$ at $Z$ and $T$ be midpoint of arc $EF$ in $AEF$ and $S$ be midpoint of arc $BC$ in $ABC$. Note that $PK$ is symmedian in $PEF$ so $\angle FPK = \angle EPN$ so we need to prove $\angle EZN = \angle EPN$ or $PZNE$ is cyclic. Note that $\angle EBP = \angle ABP = \angle ACP = \angle FCP$ and $\angle AFP = \angle AEP$ so $PEB$ and $PFC$ are similar so $P$ sends $EF$ to $BC$. Note that $A,T,S$ are collinear so $PTS$ and $PEB$ are similar and Note that $PNT$ and $PMS$ are similar so $\angle PNZ = \angle PTA = \angle PEA = \angle PEZ$ so $PZNE$ is cyclic and we're Done.