Lemma: Let $D,E,F$ be three points in the plain of $\triangle ABC$ such that $\triangle DBC,\triangle EAC,\triangle FAB$ are isosceles similar triangles then $\triangle ABC,\triangle DEF$ have the same centroid.

Back to the problem: Let $\Omega_E\cap \Omega_F=X\ ,\ \Omega_E\cap \Omega_D=Y\ ,\ \Omega_D\cap \Omega_F=Z$. Let $AX\cap \odot(BXC)=D'$ and $\angle BDC=\theta$ ; since $E,F$ are circumcenters of $AXC,AXB\Longrightarrow \angle CXD'=\angle BXD'=\frac{\theta}{2}\Longrightarrow D'C=D'B$ and $\angle BD'C=180^{\circ}-\theta$ hence $D'$ is orthocenter of $\triangle BDC$. Let $M$ be the midpoint of $BC$ , $H$ the orthocenter of $DEF$ and $G$ be the centroid of $\triangle ABC$. From the Lemma $G$ is also the centroid of $\triangle BDC$ so if $AM\cap DH=W$: $$\frac{GH}{GR}=\frac{GW}{GA}=\frac{MW-MG}{GA}=\frac{3}{2}.\frac{MW}{MA}-\frac{1}{2}=\frac{3}{2}.\frac{MD}{MD'}-\frac{1}{2}=\frac{3\cot^2(\frac{\theta}{2})}{2}-\frac{1}{2}=\text{constant}$$ Hence $R$ is fixed point which lies on $GH$. Similarly $BY$ and $CZ$ cut the Euler line of $DEF$ with the same ratio hence $R$ is radical center of $\Omega_D,\Omega_E,\Omega_F$. Q.E.D

My solution. We see triangle $ABC$ and $DEF$ are orthologic so the perpendicular through $A,B,C$ to $EF,FD,DE$, reps, are concurrent at $K$ which is the radical center of the pair of circles $ \Omega_D, \Omega_E, \Omega_F $. We shall prove that $K$ lies on Euler line of $DEF$ by using tripolar coordinates of Euler line, this is equivalent to $(DE^2-DF^2)KD^2+(EF^2-ED^2)KE^2+(FD^2-FE^2)KF^2=0$ $\iff DE^2(KD^2-KE^2)+EF^2(KE^2-KF^2)+FD^2(KF^2-KD^2)=0$ $\iff DE^2(CD^2-CE^2)+EF^2(AE^2-AF^2)+FD^2(BF^2-BD^2)=0$ $\iff DE^2(x^2-y^2)+ EF^2(y^2-z^2)+FD^2(z^2-x^2)=0$ Let $\angle DBC=\angle DCB=\angle ECA=\angle EAC=\angle FAB=\angle FBA=\theta$ and $BC=a,CA=b,AB=c,$ $DB=DC=a/(2\cos\theta)=x,EC=EA=b/(2\cos\theta)=y,FA=FB=c/(2\cos\theta)=z$ apply law of cosine, $EF^2=y^2+z^2+2yz\cos(A+2\theta)$. We need to prove that $(x^2+y^2+2xy\cos(C+2\theta))(x^2-y^2)+(y^2+z^2+2yz\cos(A+2\theta))(y^2-z^2)+(z^2+x^2+2zx\cos(B+2\theta))(z^2-x^2)=0$ $\iff xy(x^2-y^2)\cos(C+2\theta)+yz(y^2-z^2)\cos(A+2\theta)+zx(z^2-x^2)\cos(B+2\theta)=0$ $\iff ab(a^2-b^2)\frac{\cos(C+2\theta)}{\cos^4\theta}+bc(b^2-c^2)\frac{\cos(A+2\theta)}{\cos^4\theta}+ca(c^2-a^2)\frac{\cos(C+2\theta)}{\cos^4\theta}=0$ $\iff ab(a^2-b^2)(\cos C\cos 2\theta-\sin C\sin 2\theta)+bc(b^2-c^2)(\cos A\cos 2\theta-\sin A\sin 2\theta)+ca(c^2-a^2)(\cos B\cos 2\theta-\sin B\sin 2\theta)=0$ $\iff \cos 2\theta((a^2-b^2)(a^2+b^2-c^2)+(b^2-c^2)(b^2+c^2-a^2)+(c^2-a^2)(c^2+a^2-b^2))-\sin 2\theta(\frac{abc}{R}(a^2-b^2+b^2-c^2+c^2-a^2))=0$ The last equality is true, we are done.

My proof: Lemma 1: Rt$\triangle ABC$$\overset{-}{\sim}$ Rt$\triangle ADE$. $AB,AD$ are hypotenuses.$M$ is the midpoint of $BD$,then $M$ is on the perpendicular bisector of $CE$. Proof for lemma 1: Let $F,G$ be the midpoint of $AB,AD$, respectively. $MF= \frac{1}{2} DA=EG,MG=\frac{1}{2} BA=CF,\measuredangle (MG,CF)=2 \measuredangle ABC=2\measuredangle AED=\measuredangle (GE,MF)$ $\Longrightarrow \triangle MCF \overset{+}{\sim} \triangle EMG \Longrightarrow MC=ME.$ Lemma 2(Obviously): $\triangle ABC$ and $\triangle A_1B_1C_1$ are both orthologic with $\triangle DEF$.Let $P$ be the orthologic center of $ \triangle ABC$ WRT $\triangle DEF$,$P_1$ be the orthologic center of $\triangle A_1B_1C_1$ WRT $\triangle DEF$. Let $A_2\in AA_1,B_2\in BB_1,C_2\in CC_1$ such that $\frac{AA_2}{AA_1}=\frac{BB_2}{BB_1}=\frac{CC_2}{CC_1}.$ Then $\triangle A_2B_2C_2$ and $\triangle DEF$ are orthologic, the orthologic center $P_2$ of $\triangle A_2B_2C_2$ WRT $\triangle DEF$ is on $PP_1,$ meanwhile $\frac{PP_2}{P_1P_2} = \frac{AA_2}{A_1A_2}$. Lemma 3: the same as the lemma mentioned in post #2. Back to the problem: Define $O,G$ as the circumcenter and centroid of $\triangle DEF$,respectively. Let $E_A \in AC,F_A \in AB$ such that $E_AE \perp AE,F_AF \perp AF$,$A_2$ be the midpoint of $E_AF_A$,similarly define $B_2,C_2$.From lemma 1 we can know $A_2$ is on the perpendicular bisector of $EF$,similarly for $B_2,C_2.$ Hence $\triangle A_2B_2C_2$ and $\triangle DEF$ are orthologic, $O$ is the the orthologic center of $\triangle A_2B_2C_2$ WRT $\triangle DEF$. Let $P$ be the orthologic center of $\triangle ABC$ WRT $\triangle DEF$(scilicet the radical center of $\Omega_D, \Omega_E, \Omega_F$),$P_1$ be the orthologic center of $\triangle A_1B_1C_1$ WRT $\triangle DEF$. Obviously $A_2\in AA_1,B_2\in BB_1,C_2\in CC_1$,meanwhile $\frac{AA_2}{AA_1}=\frac{BB_2}{BB_1}=\frac{CC_2}{CC_1}.$ From lemma 2 we can know $O,P,P_1,$ are colinear,and obviouly $P,P_1,G$ are colinear $\Longrightarrow $$P,P_1 \in $ the Euler line of $\triangle DEF.$

Official solution : Let $ T $ be the second intersection of $ \Omega_E, \Omega_F $ and let $ H_D $ be the second intersection of $ AT $ with $ \odot (BTC). $ From $$ \measuredangle BCH_D = \measuredangle BTA = \tfrac{1}{2} \measuredangle BFA = \tfrac{1}{2} \measuredangle CDB $$we get $ BH_D \perp CD. $ Similarly, $ CH_D \perp BD, $ so $ H_D $ is the orthocenter of $ \triangle BDC. $ Analogously, if $ H_E, H_F $ is the orthocenter of $ \triangle CEA, \triangle AFB, $ respectively, $ BH_E, CH_F $ is the radical axis of $ (\Omega_F, \Omega_D), $ $ (\Omega_D, \Omega_E), $ respectively, so the radical center $ P $ of $ \Omega_D, \Omega_E, \Omega_F $ lies on $ AH_D, BH_E, CH_F. $ Let $ X, Y, Z $ be the reflection of $ D, E, F $ in $ BC, CA, AB, $ respectively. From $ \triangle BXC \stackrel{+}{\sim} \triangle BFA $ $ \Longrightarrow $ $ \triangle BCA \stackrel{+}{\sim} \triangle BXF, $ so $ \tfrac{AE}{BF} = \tfrac{CA}{AB} = \tfrac{FX}{BF} $ $ \Longrightarrow $ $ AE = FX. $ Similarly, we get $ AF $ $ = $ $ EX, $ so $ AEXF $ is a parallelogram. Let $ O_T $ be the circumcenter of $ \triangle DEF $ and let $ Q $ be the reflection of $ P $ in $ O_T, $ then from the discussion above we get $ XQ \perp EF. $ Analogously, $ YQ, ZQ $ is perpendicular to $ FD, DE, $ respectively. Let $ H_T $ be the orthocenter of $ \triangle DEF. $ Notice $ \tfrac{DX}{DH_D} = \tfrac{EY}{EH_E} = \tfrac{FZ}{FH_F}, $ so $ H_T, P, Q $ are collinear. i.e. $ P $ lies on the Euler line $ O_TH_T $ of $ \triangle DEF. $ $ \blacksquare $

$\textbf{Proof :}$ $\textbf{Lemma :}$ Given a $ \triangle ABC $ and three points $ D, E, F $ such that $ DB = DC, $ $ EC = EA, $ $ FA = FB, $ $ \measuredangle BDC = \measuredangle CEA = \measuredangle AFB$. Let $D'$ be reflection of $D$ wrt $BC$ and $D_1$ be the circumcenter of $(D'BC)$. Then $D_1$ lies on the perpendicular bisector of $EF$. Now as we know triangles $ABC$ and $DEF$ have same centroid, so it is enough to prove that the lines through the midpoints of $BC, CA, AB$ which are perpendicular to sides of $DEF$ meet at point $X$ on the Euler line of $DEF$. Consider point $D_1$ as in lemma. Similarly define $E_1, F_1$. Let $D'$ be midpoint of $BC$, similarly define $E', F'$. We get that $\frac{DD'}{DD_1} = \frac{EE'}{EE_1} = \frac{FF'}{FF_1}$, so $X$ lie on the line between the orthocenter of $DEF$ and the circumcenter of $DEF$. Done

andria wrote: Lemma: Let $D,E,F$ be three points in the plain of $\triangle ABC$ such that $\triangle DBC,\triangle EAC,\triangle FAB$ are isosceles similar triangles then $\triangle ABC,\triangle DEF$ have the same centroid.

Back to the problem: Let $\Omega_E\cap \Omega_F=X\ ,\ \Omega_E\cap \Omega_D=Y\ ,\ \Omega_D\cap \Omega_F=Z$. Let $AX\cap \odot(BXC)=D'$ and $\angle BDC=\theta$ ; since $E,F$ are circumcenters of $AXC,AXB\Longrightarrow \angle CXD'=\angle BXD'=\frac{\theta}{2}\Longrightarrow D'C=D'B$ and $\angle BD'C=180^{\circ}-\theta$ hence $D'$ is orthocenter of $\triangle BDC$. Let $M$ be the midpoint of $BC$ , $H$ the orthocenter of $DEF$ and $G$ be the centroid of $\triangle ABC$. From the Lemma $G$ is also the centroid of $\triangle BDC$ so if $AM\cap DH=W$: $$\frac{GH}{GR}=\frac{GW}{GA}=\frac{MW-MG}{GA}=\frac{3}{2}.\frac{MW}{MA}-\frac{1}{2}=\frac{3}{2}.\frac{MD}{MD'}-\frac{1}{2}=\frac{3\cot^2(\frac{\theta}{2})}{2}-\frac{1}{2}=\text{constant}$$ Hence $R$ is fixed point which lies on $GH$. Similarly $BY$ and $CZ$ cut the Euler line of $DEF$ with the same ratio hence $R$ is radical center of $\Omega_D,\Omega_E,\Omega_F$. Q.E.D I think the diagram is wrong... $AERF$ in diagram doesn't make parallelogram

Here is a Different Proof. TelvCohl wrote: Given a $ \triangle ABC $ and three points $ D, E, F $ such that $ DB = DC, $ $ EC = EA, $ $ FA = FB, $ $ \measuredangle BDC = \measuredangle CEA = \measuredangle AFB. $ Let $ \Omega_D $ be the circle with center $ D $ passing through $ B, C $ and similarly for $ \Omega_E, \Omega_F. $ Prove that the radical center of $ \Omega_D, \Omega_E, \Omega_F $ lies on the Euler line of $ \triangle DEF. $ Proposed by Telv Cohl $\textbf{LEMMA 1:-}(\text{Well Known})$ $ABC$ be a triangle and $\Delta DEF$ be a Kiepert Triangle of $\Delta ABC$. Then $\Delta ABC$ and $\Delta DEF$ have the same centroid $G$. Proof:- Notice that $\Delta BDC\stackrel{-}{\sim}\Delta CEA\stackrel{-}{\sim}\Delta AFB$. Let $D^*$ be the reflection of $D$ over $BC$. Then $\Delta CD^*B\stackrel{+}{\sim}\Delta CEA\implies\Delta CED^*\stackrel{+}{\sim}\Delta CAB$. So, $\measuredangle FAE=A+2\measuredangle CAE$ also $\measuredangle AED^*=\pi-2\measuredangle CAE-A\implies\overline{AF}\|\overline{D^*E}$. Similarly we get $\overline{AE}\|\overline{D^*F}$. Hence, $D^*$ is the reflection of $A$ over midpoint of $EF$. Let $\{T,K\}$ be the midpoints of $BC,EF$ repectively, So clearly if $\overline{DK}\cap\overline{AT}=G$. Then $\frac{GD}{GK}=\frac{GA}{GT}=2\implies\{G\}$ is the Centroid of $\Delta ABC$ and $\Delta DEF$. $\qquad$ $\blacksquare$ $\textbf{LEMMA 2:-}$ Given two (not homothetic) triangles $\Delta ABC $ and $\Delta DEF $. Let $ P $ be a point such that the line passing through $ D, $ $ E, $ $ F $ and parallel to $\overline{AP}, $ $\overline{BP}, $ $\overline{CP} $, respectively are concurrent. Then $ P $ lies on a circumconic of $ \triangle ABC $ or the line at infinity. Proof:- See Lemma 1 here(#9) $\qquad$ $\blacksquare$ $\textbf{LEMMA 3:-}$. $ABC$ be a triangle and $\Delta DEF$ be a Kiepert Triangle of $\Delta ABC$. If $\{H_A\}$ is the orthocenter of $\Delta BDC$. Then $\overline{AH_A}\perp\overline{EF}$. Proof- Notice that $\{\Delta DEF,\Delta ABC\}$ are Orthologic Triangles. Let $X$ be the Orthology Center of $\Delta ABC,\Delta DEF$ except $O$ where $O$ is the circumcenter of $\Delta ABC$. Let $Y$ be the Perspector of $\{\Delta ABC,\Delta DEF\}$. Then from $\textbf{LEMMA 2}$ we get that $\{A,B,C,X,Y\}$ lie on a Conic and it's well known that $X\in\mathcal{K}$ which is the Kiepert Hyperbola of $\Delta ABC$. Let $\measuredangle DBC=\theta$. Now by Sondat's Theorem we get that $O,X,Y$ are collinear. So from (2) here(#5) we get that $Y$ is $K(90^\circ-\theta)$ of $\Delta ABC\implies\overline{AY}$ passes through $H_A$. Hence, $\overline{AH_A}\perp\overline{EF}$. $\qquad$ $\blacksquare$ ____________________________________________________________________________________________ Coming back to the Problem. Let $H$ be the Orthocenter of $\Delta DEF$. So from $\textbf{LEMMA 1}$ we get that $\overline{HG}$ is the Euler Line $(\mathcal E)$ of $\Delta DEF$ and define $\{H_B,H_C\}$ as the Orthocenters of $\Delta ACE,ABF$ respectively. Then $\Delta BDC\cup H_A\stackrel{-}{\sim}\Delta CEA\cup H_B\stackrel{-}{\sim}\Delta AFB\cup H_C (\bigstar)$. Let $\overline{HD}\cap\overline{AG}=\{V\}$ and $\overline{AH_A}\cap\mathcal E=\{U\}$. So, $\frac{GV}{GA}=\frac{GT-TV}{\frac{2}{3}TA}=\frac{1}{2}-\frac{TV}{TA}=\frac{1}{2}-\frac{TD}{TH_A}$ and clearly this is a Constant $\mathcal P$ from $(\bigstar)$. Also clearly the Radical Center of $\Omega_D,\Omega_E,\Omega_F$ is the Orthology Center $X$ of $\Delta ABC$. So, $\overline{AH_A}\cap\overline{BH_B}\cap\overline{CH_C}=X\in\mathcal{E}$. $\qquad$ $\blacksquare$ Remark:- Here goes a more General Result, which I have not proved yet. GENERALIZATION:- $ABC$ be a triangle and $P$ be an arbitary point on the Kiepert Hyperbola $\mathcal K$ of $\Delta ABC$. Let $\Delta DEF$ be a Kiepert Triangle of $\Delta ABC$. Then the Parallel Lines from $D,E,F$ to $\overline{AP},\overline{BP},\overline{CP}$ councurs at a point $Q$ and also $Q$ lies on the Kiepert Hyperbola $\mathcal{K}_1$ of $\Delta DEF$ and $\overline{PQ}$ passes through the Centroid $G$ of $\Delta ABC$.