+Let $I_1;I_2;I_3$ be the excenter of $\triangle ABC$ respectively. $G_1;G_2$ be the centroid of $\triangle DEF;\triangle I_1I_2I_3$.$O_1;O_2;O_2$ be the circumcenter of $\triangle DEF ;\triangle ABC ; \triangle I_1I_2I_3$ +We'll have $\triangle DEF;\triangle I_1I_2I_3$ is homothetic.Let $G_3$ be the center of a homothety takes $\triangle DEF$ to $\triangle I_1I_2I_3$.We'll get $\overline {G_1;G_2;G_3} ;\overline {G_3;O_1;O_3} ; \overline {O_1;O_2;O_3}$.So We'll conclude $\overline{G_1;O_1;O_2}$ Our proof is completed Our proof is completed *Remark : Notation $\overline {A;B;C}$ mean $A;B;C$ are collinear

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Let $O_1,I,O_2$ be circumcenter of $\Delta ABC,\Delta DEF,\Delta XYZ$ respectively. Let $A'=AI\cap (O_1),B'=BI\cap (O_1),C'=CI\cap (O_1)$ It's easy to prove that $XZ\parallel FD\parallel C'A'\Rightarrow \frac{IX}{IA'}=\frac{IY}{IB'}$ Smilarity we have $\frac{IY}{IB'}=\frac{IZ}{IC'}$ so $I$ is the homothetic center of two triangles $XYZ$ and $A'B'C'$ Thus $O_1,I,O_2$ are collinear.

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Let $I$ and $\triangle I_AI_BI_C$ be the incenter and excentral triangle of $\triangle ABC$. Also, Let $V$ be the Bevan point of $\triangle ABC$. Then $V$ lies on line $OI$, where $O$ is the circumcenter of $\triangle ABC$. Now, $\angle FDB=\angle FIB=90^{\circ}-\frac{B}{2}=\angle I_ABD \Rightarrow DF \parallel I_AI_C$. Thus, we get that $\triangle DEF$ and $\triangle I_AI_BI_C$ are homothetic, and their homothety center lies on the line joining their centers, i.e. line $IV$. This gives that line $OI$ is the Euler line of $\triangle DEF$, proving that the nine point center of $\triangle DEF$ (i.e. the circumcenter of $\triangle XYZ$) lies on line $OI$.

Amir Hossein wrote: The circle inscribed in a triangle $ABC$ touches the sides $BC,CA,AB$ in $D,E, F$, respectively, and $X, Y,Z$ are the midpoints of $EF, FD,DE$, respectively. Prove that the centers of the inscribed circle and of the circles around $XYZ$ and $ABC$ are collinear. It is very easy with inversion around the incircle of the triangle $ABC$. Then the circumcircle of the triangle $ABC$ will be the nine points circle of $DEF$ (the nine points circle of $DEF$ is the circumcircle of the triangle $XYZ$). So we have $O$, $I$ and the circumcenter of $XYZ$ are collinear. So we also have $OI$ is the euler line of the contact triangle of the triangle $ABC$.