I suppose there is a solution possible using complex geometry. In my configuration $\triangle ABC $ is anticlockwise and $T,D $ lie on the same side of $EF $ which is of course opposite to $A $. Now set $\odot (ABC) $ as the complex unit circle and let the complex coordinate of any point $X $ be $x $ unless stated otherwise. $a'+a=0$ is immediate. As for $D $, $\angle(CD,BD)=\tfrac{\pi}{3} $. Setting $\omega=e^{\tfrac {-i\pi}{3}}$, we see that $\omega^3=1,\overline\omega=\omega^2$. Then for my configuration, $(d-c)(-\omega)=(d-b)\implies d=-b\omega-c\omega^2\implies \bar d=\frac {d}{bc}$ using $b\bar b=c\bar c=1,\overline\omega=\omega^2$. Clearly the equation for the line $l\perp $ to $A'D $ at $A'$ is $\frac {z+a}{d+a}+\frac {\bar z+\bar a}{\bar d+\bar a}=0\implies \frac {z+a}{d+a}+\frac {abc\bar z+bc}{ad+bc}=0$. Now the equation for line $AB $ is $z+\bar zab-(a+b)=0\implies \bar z=\frac {(a+b)-z}{ab} $. Now putting this value of $\bar z $ in $l $ will give the coordinate of $l\cap AB\equiv F=f $. Solving for this coordinate gives $f=\frac {a (ad+bc)+c (a+2b)(a+d)}{(c-a)d+(a-b)c}$ which then gives $\bar f=\frac {2bc (d+a)+(2b+a)(bc+ad)}{2ab\left [d (a-c)-c (a-b)\right]} $. Interchanging $b $ and $c $ in the expressions of $f,\bar f $ gives $e,\bar e $. So we get $e=\frac {a (ad+bc)+b (a+2c)(a+d)}{(b-a)d+(a-c)b} $, $\bar e=\frac {2bc (a+d)+(2c+a)(bc+ad)}{2ac\left [d (a-b)-b (a-c)\right]} $. Now we find the coordinate of $t $. For my configuration, $\angle(TE,TF)=\frac {2\pi}{3} $, so $(t-e)\omega^2=(t-f) \implies t=\frac {f\omega-e}{\omega-1},\bar t=\frac {\bar e\omega-\bar f}{\omega-1} $. Now the coordinate of the Nine Point Centre of $\triangle ABC $ is $\frac {a+b+c}{2} $, hence we need that $\frac {t-a}{\tfrac {a+b+c}{2}-a}\in\mathbb {R}\iff \frac {t-a}{b+c-a}\in\mathbb {R}\iff\frac {t-a}{b+c-a}=\frac {\bar t-\bar a}{\bar b+\bar c-\bar a} $. Using the expressions for $t,\bar t $ this reduces to showing that $\left (\frac {1}{b}+\frac {1}{c}-\frac {1}{a}\right)(f\omega-e-a (\omega-1))=(b+c-a)\left (\bar e\omega-\bar f-\frac {(\omega-1)}{a}\right) $. Using the already found values of $e,\bar e,f,\bar f,d$, the above required equality turns into an identity hence we are done. NOTE : Since I am on mobile (and I am lazy) it is very inconvenient to type out the calculations in the last step, I leave it to the reader to verify. I have worked it out by hand on paper and the calculations seem(?!) to be fine. However please do check all the calculations and notify me if anything seems incorrect. And is it just me or is the problem really workable using complex geometry, because the actual bash took me just $50$ minutes to complete.

Nice problem! Congratulations! This problem actually begins from this easy lemma, Lemma : Given an equilateral triangle $ABC$ with center $O$, $D$ is a point on ray $AO$ and $E$ is the reflection of $D$ through $BC$. Then $\overline{OE}$ $\cdot $ $\overline{OA}$ $=$ $\text{Power}(O,(D,BC))$. Proof : Let $F$ be the projection of $A$ onto $BC$. Note that $BF^2$ $=$ $3OF^2$ $=$ $OF^2$ $+$ $OA$ $\cdot$ $(OE$ $+$ $EF)$ hence $OE$ $\cdot $ $OA$ $=$ $BF^2$ $-$ $OF^2$ $-$ $2OF$ $\cdot $ $EF$ $=$ $BE^2$ $-$ $EF^2$ $-$ $OF^2$ $-$ $2OF$ $\cdot $ $EF$ $=$ $BE^2$ $-$ $OD^2$ $=$ $\text{Power}(O,(D,BC))$. $\square$ _________________________________ Back to main problem : Let $DA'$ meets $\Gamma $ at the second point $G$, $G'$ is the reflection of $G$ through the perpendicular bisector of $BC$. $O$ is the circumcentre of $\triangle ABC$ and $O'$ is the reflection of $O$ through $BC$. $AO'$ meets $\Gamma $ at the second point $Z$, $GZ$ meets $OD$ at $S$. Note that $\angle DG'Z$ $=$ $\angle DO'Z$ so $G'$, $O'$, $Z$, $D$ lies on a circle. Hence, from the lemma we obtain that $S$ is the center of $\triangle DBC$, then $\triangle GCB \cup S$ $\sim $ $\triangle AEF \cup T$. This implies that $\angle TAB$ $=$ $\angle SGC$ $=$ $\angle O'AB$, thus $AT$ passes through the 9 - point center of $\triangle ABC$. $\square$ PS. I think this problem can be generalized by generalize the lemma.

Tatsuya wrote: Nice problem! Congratulations! This problem actually begins from this easy lemma, Lemma : Given an equilateral triangle $ABC$ with center $O$, $D$ is a point on ray $AO$ and $E$ is the reflection of $D$ through $BC$. Then $\overline{OE}$ $\cdot $ $\overline{OA}$ $=$ $\text{Power}(O,(D,BC))$. Proof : Let $F$ be the projection of $A$ onto $BC$. Note that $BF^2$ $=$ $3OF^2$ $=$ $OF^2$ $+$ $OA$ $\cdot$ $(OE$ $+$ $EF)$ hence $OE$ $\cdot $ $OA$ $=$ $BF^2$ $-$ $OF^2$ $-$ $2OF$ $\cdot $ $EF$ $=$ $BE^2$ $-$ $EF^2$ $-$ $OF^2$ $-$ $2OF$ $\cdot $ $EF$ $=$ $BE^2$ $-$ $OD^2$ $=$ $\text{Power}(O,(D,BC))$. $\square$ _________________________________ Back to main problem : Let $DA'$ meets $\Gamma $ at the second point $G$, $G'$ is the reflection of $G$ through the perpendicular bisector of $BC$. $O$ is the circumcentre of $\triangle ABC$ and $O'$ is the reflection of $O$ through $BC$. $AO'$ meets $\Gamma $ at the second point $Z$, $GZ$ meets $OD$ at $S$. Note that $\angle DG'Z$ $=$ $\angle DO'Z$ so $G'$, $O'$, $Z$, $D$ lies on a circle. Hence, from the lemma we obtain that $S$ is the center of $\triangle DBC$, then $\triangle GCB \cup S$ $\sim $ $\triangle AEF \cup T$. This implies that $\angle TAB$ $=$ $\angle SGC$ $=$ $\angle O'AB$, thus $AT$ passes through the 9 - point center of $\triangle ABC$. $\square$ PS. I think this problem can be generalized by generalize the lemma. Nice

Official solution : Lemma 1 : Let $ P, Q $ be the isogonal conjugate WRT $ \triangle ABC $ and $ O, O_a, O_b, O_c, T $ be the circumcenter of $ \triangle ABC, $ $ \triangle BPC, $ $ \triangle CPA, $ $ \triangle APB, $ $ \triangle O_aO_bO_c, $ respectively. Then $ PQ \parallel OT. $ Proof : Let $ \triangle Q_aQ_bQ_c $ be the pedal triangle of $ Q $ WRT $ \triangle ABC $ and $ V $ be the circumcenter of $ \triangle Q_aQ_bQ_c. $ It is well-known that $ V $ is the midpoint of $ PQ, $ so notice $ \triangle Q_aQ_bQ_c \cup Q \cup V $ and $ \triangle O_aO_bO_c \cup O \cup T $ are homothetic we get $ PQ \parallel OT. $ $ \blacksquare $ ______________________________ Lemma 2 : Given an equilateral triangle $ \triangle BUC $ and a point $ A. $ Let $ V $ be the isogonal conjugate of $ U $ WRT $ \triangle ABC. $ Then $ UV $ is parallel to the Euler line of $ \triangle ABC. $ Proof : Let $ BV, CV $ cut the $ \odot (ABC) $ at $ E F, $ respectively. Clearly, $ \triangle AEF $ is an equilateral triangle, so the line connecting the midpoint $ M, N $ of $ BC, EF $ is parallel to the Euler line of $ \triangle ABC. $ Note that $ \measuredangle VBU = \measuredangle FAC, $ $ \measuredangle VCU = \measuredangle EAB, $ so $$ [\triangle UMV] - [\triangle UNV] = \tfrac{1}{2} \left( [\triangle UBV] + [\triangle UCV] \right ) - \tfrac{1}{2} \left( [\triangle UEV] + [\triangle UFV] \right) = \tfrac{1}{2} \left( [\triangle UBE] + [\triangle UCF] \right) = 0, $$hence we conclude that $ UV $ $ \parallel $ $ MN $ $ \Longrightarrow $ $ UV $ is parallel to the Euler line of $ \triangle ABC. $ $ \blacksquare $ ______________________________ Back to the main problem Let $ X $ be the circumcenter of $ \triangle ABC $ and $ V $ be the reflection of $ X $ in $ BC. $ Since the 9-point center of $ \triangle A'BC $ is the midpoint of $ A'V, $ so it suffices to prove $ AT $ is parallel to the Euler line of $ \triangle A'BC. $ Let $ O, $ $ Y, $ $ Z, $ $ K $ be the circumcenter of $ \triangle BCD, \triangle A'BD, \triangle A'CD, \triangle OYZ, $ respectively. Clearly, $ \triangle KYZ $ is an isosceles triangle with base $ YZ $ and base angle $ 30^{\circ}, $ so notice $ YZ, ZX, XY $ is perpendicular to $ A'D, A'C, A'B, $ resp. we get $ \triangle AEF $ and $ \triangle XZY $ are homothetic, hence $ \triangle AEF \cup T $ and $ \triangle XZY \cup K $ are homothetic $ \Longrightarrow $ $ AT \parallel XK. $ Let $ D' $ be the isogonal conjugate of $ D $ WRT $ \triangle A'BC, $ then by Lemma 1 we get $ DD' \parallel XK, $ hence note that $ DD' $ is parallel to the Euler line of $ \triangle A'BC $ (Lemma 2) we conclude that $ AT $ and the Euler line of $ \triangle A'BC $ are parallel. $ \blacksquare $ ____________________________________________________________ Remark : 1. Corollary of this problem : Given an equilateral triangle $ \triangle ABC $ and a point $ P. $ Then the Euler line of $ \triangle BPC, $ $ \triangle CPA, $ $ \triangle APB $ are concurrent. 2. Generalization of Lemma 2 : Given a $ \triangle ABC $ with circumcenter $ O, $ orthocenter $ H. $ Let $ U $ be a point such that $ \triangle BUC $ is an isosceles triangle with base $ BC $ and base angle $ \theta. $ Let $ V $ be the isogonal conjugate of $ U $ WRT $ \triangle ABC $ and $ P $ be the intersection of $ OH, $ $ UV. $ Then $$ \frac{OP}{OH} = \frac{1}{4 \cos^2 \theta - 1} \ . $$

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1. Corollary of this problem : Given an equilateral triangle $ \triangle ABC $ and a point $ P. $ Then the Euler line of $ \triangle BPC, $ $ \triangle CPA, $ $ \triangle APB $ are concurrent. Not only the OH lines (Euler lines). Also the OI, OK and other lines passing through O (but not all ...) See https://beta.groups.yahoo.com/neo/groups/Hyacinthos/conversations/messages/23296

Very nice problem! Here is my solution. Extend $A'D$ to meet $\Gamma$ at $P$. Then $AP\perp A'D\perp EF$ thus $\measuredangle PBC = \measuredangle PAC = \measuredangle EFA$. Hence $\triangle PBC\stackrel{-}{\sim} \triangle AFE$. Now let $Q$ be the center of $\triangle BDC$. Hence $\triangle PBC\cup Q\stackrel{-}{\sim}\triangle AFE\cup T$. Let $H, O$ be orthocenter, circumcenter $\triangle A'BC$. Since $\measuredangle(PQ, BC)=-\measuredangle(AT, EF)$, we get \begin{align*} \measuredangle(PQ, A'D) + \measuredangle(AT, A'H) &=\measuredangle(PQ, A'H) + \measuredangle(AT, A'D) \\ &= \measuredangle(PQ, BC) + \measuredangle(AT, EF) \\ &= 0 \end{align*}Hence we will be done if we show that $\measuredangle QPA' = \measuredangle(OH, A'H)$ or $\measuredangle QPA' = \measuredangle OHA'$. Seeing this, construct point $K$ such that $A'KOH$ is parallelogram. Hence the angle relation becomes to $\measuredangle QPA' = \measuredangle QKA'$ or $Q, A', P, K$ are concyclic. To prove this fact, we simply length bash. Let $M$ be the midpoint of $BC$ and $OM=x$. Hence $OK=A'H=2x$. Let $QM = y$. We see $BC = \sqrt{3}y$ and $QD = 2y$. Hence \begin{align*} DA'\cdot DP &= OD^2 - R^2 = (x+3y)^2 - (x^2+3y^2) = 6xy + 6y^2 \\ DQ\cdot DK &= (2x)(3x+3y) = 6xy+6y^2\\ \end{align*}both are equal so we are done.

Here's very strange solution inspired by China TST 2012 Test3 1. Suppose now that triangle $ABC$ is acute. Let $O,H$ be the circumcenter, orthocenter of triangle $ABC$ and $M$ be the midpoint of side $BC$. Define $P,Q,K,L$ such that $P,Q$ lie on $CA,AB$ and triangles $BKH,HPQ,CHL$ are equilateral as in the picture. Also, the second intersection of $(KHP),(LHQ)$ be $D'$. By spiral similarity, $\triangle BHQ\sim\triangle KHP$ and $\triangle CHP\sim\triangle LHQ$ thus $\angle HD'P=\angle HD'Q=90^\circ-\angle A$ also we have $D'P=D'Q\Rightarrow\angle D'HP=\angle QHD'=30^\circ$. Moreover, by simple angle chasing it holds that $$\angle HKD'=\angle D'LH=120^\circ-\angle A, \angle KD'L=\angle LHK=60^\circ+\angle A$$means that $KD'LH$ is a parallelogram. From the above, we get $BH=BK=D'T, CH=CL=D'S$ and $\angle BHC=\angle BKD'=\angle D'LC=180^\circ-\angle A$ so $\triangle BKD'\equiv\triangle BHC\equiv\triangle D'LC$ and $BCD'$ is a equilateral triangle. Let $G$ be the circumcenter of triangle $HPQ$ and $O'$ be the reflection of $O$ wrt $BC$. From $(A',H)$ and $(D,D')$ are symmetry wrt $M$, we get $A'D\parallel HD'$. Moreover, $PQ\perp HD'\Rightarrow PQ\parallel EF$ hence $APGQ$ and $AEFT$ are similar. Now it suffices to show that $A,G,O'$ are collinear. Notice that $\angle D'PQ+\angle O'BC=90^\circ,AH=2O'M$ and $AH\parallel D'O'$ we have the desired conclusion from $$\displaystyle\frac{HG}{HD'}=\frac{2\sqrt{3}}{3\sqrt{3}+3\tan\angle D'PQ}=\frac{2\tan\angle O'BC}{2\tan\angle O'BC+(\tan\angle O'BC+\sqrt{3})}=\frac{AH}{AH+O'D'}.$$

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Solved with hakN, BarisKoyuncu, Cookierokie, InfinityFun. $GEF$ is an equilateral triangle and $T$ is the center of it. $\textbf{Claim:}$ $AG \perp BC$. Proof: $S \in AG$ with $AEFS$ is cyclic. Note that $\angle FSE= \angle BA'C$ and $\angle FSG=180- \angle A'EC=\angle DA'C$. Thus $BCD \cup A' \cup J=EFG \cup S \cup A$ which implies $\angle FAS=\angle A'JC=\angle A'AC$ as desired. See the lemma here: https://artofproblemsolving.com/community/c6h2855130 Applying the lemma we get $$\frac{d(T,AA')}{d(T,AH)}\overset{lemma}=2 \cos A=\frac{AH}{AO}\overset{trig-ratio-lemma}=\frac{\sin N_9AO}{\sin HAN_9}=\frac{d(N_9,AA')}{d(T,AH)}$$which implies $A,N_9,T$ are collinear.