Found this https://artofproblemsolving.com/community/c6h1184638p5748733

It was my proposal... didn't expect it to be created by someone else before >_<

any solution

My original solution is as follows: First we prove that the circles $(DQR), (EPR), (FPQ)$ are all tangent to $(DEF)$. Denote $A_1, B_1, C_1$ be the midpoints of $BC, CA, AB$ respectively, then evidently $A_1, B_1, C_1$ lie on $(DEF)$, which is the nine-point circle of $\triangle ABC$. Observe $\angle B_1C_1C = \angle C_1CB = \angle RQB = \angle RQB_1$, implying $C_1B_1QR$ is cyclic. So, applying Radical Axis Theorem on circles $(RC_1B_1Q), (ABC), (AB_1C_1)$, then $K = QR \cap B1C1$ is the radical center of these $3$ circles. However we know that $(ABC), (AB_1C_1)$ are tangent, thus $K$ which lies on their radical axis must also lie on their tangent, implying $KA$ is tangent to both $(ABC), (AB_1C_1)$. Observe $D$ is the reflection of $A$ about $B_1C_1$, thus $KD = KA$. Finally this gives $KD^2 = KA^2 = KC_1 \cdot KB_1 = KR\cdot KQ$, thus $KD$ is tangent to both $(DQR)$ and $(DB_1C_1)$. This proves our claim that $(DQR), (DEF) = (DB1C1)$ are tangent, and likewise we obtain $(EPR), (FPQ)$ are both tangent to $(DEF)$ as well. Denote $X, Y, Z$ be the pole of $EF, FD, DE$ in circle $(DEF)$. Observe $XE, XF$ are the radical axis of circles $(EPR), (DEF)$ and $(FPQ), (DEF)$ respectively, so $X$ must also lie on the radical axis of $(EPR)$ and $(FPQ)$. So $PX$ is the radical axis of $(EPR)$ and $(FPQ)$. Repeating this argument to the other two pair of circles, we see that it is equivalent to prove that $PX, QY, RZ$ are concurrent at the Euler Line. We will show that $QY \cap RZ$ lies on the Euler Line, then repeating for the other two pair of lines will solve the problem. Recall that $QR \cap B_1C_1 = K$ lies on the tangent line of $(DEF)$ at $D$, which is line $YZ$. In other words, $YZ, B_1C_1, QR$ are concurrent at $K$, thus by Desargues Theorem, $\triangle YQB_1$ and $\triangle ZRC_1$ are line perspective too. Consequently, $QY \cap RZ, G = QB_1 \cap RC_1$ and $YB_1 \cap ZC_1$ are colinear. However it is obvious that $YB_1$ is the perpendicular bisector of $DF$, thus $YB_1$ contains the point $N_9$, the nine point center of $\triangle ABC$, which is the center of the circle $(DEF)$. Similarly, $N_9 \in ZC_1$ as well, so $YB_1 \cap ZC_1 = N_9$. We conclude that $QY \cap RZ$, lies on the line $GN_9$, which is the Euler Line of $\triangle ABC$, and we are done.