Let $\vec{v}=\vec{AA'}$. Let $\nu$ and $\tau$ be the translation mappings given by $\vec{v}$ and $\frac{-\vec{v}}{2}$ respectively. Let $B''=\nu(B), C''=\nu(C), A''=\nu(A)=A'$. Let $A_{1}, B_{1}, C_{1}$ be the midpoints of $AA', BB' ,CC'$ and $A_{2}, B_{2}, C_{2}$ be the midpoints of $A''A', B''B', C''C'$. Then $\vec{A_{1}A_{2}} = \frac{\vec{AA''}}{2} = \frac{\vec{v}}{2},\ \vec{B_{1}B_{2}} = \frac{\vec{BB''}}{2} = \frac{\vec{v}}{2},\ \vec{C_{1}C_{2}} = \frac{\vec{CC''}}{2} = \frac{\vec{v}}{2}$. Then $A_{1}=\tau(A_{2}), B_{1}=\tau(B_{2}), C_{1}=\tau(C_{2})$. If $\triangle{ABC}$ and $\triangle{A'B'C'}$ have the same orientation, then so do $\triangle{A''B''C''}$ and $\triangle{A'B'C'}$. Then $B''B' $ is the image of $C''C'$ under a $\frac{\pi}{3}$ rotation with center at $A'=A''=A_{2}$ (clockwise or counterclockwise). Then $B_{2}$ is the image of $C_{2}$ under the same rotation; then $\triangle{A_{2}B_{2}C_{2}} $ is equilateral; and then $\triangle{A_{1}B_{1}C_{1}}$ is also equilateral (because is the image under $\tau$). If $\triangle{ABC}$ and $\triangle{A'B'C'}$ have different orientation, then so do $\triangle{A''B''C''}$ and $\triangle{A'B'C'}$. Then $A'', B'', C''$ are the images of $A', B', C'$ under a reflexion with respect to a line $l$. Then $A_{2}, B_{2}, C_{2}$ lie on $l$, then they are collinear and then $A_{1}, B_{1}, C_{1}$ are also collinear.