Similarly $M_2 = \frac{1}{3}\left(\frac{C+A}{2} + E + Q\right)$, $M_3 = \frac{1}{3}\left(\frac{A+B}{2} + F + R\right)$.

Now, consider $\omega = \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}$, and note that $\omega^3 = 1$ and $1+\omega+\omega^2=0$. By the given conditions, $\triangle DEF$ is an equilateral triangle, so $D+\omega E+\omega^2 F = 0$. Also, $\triangle PD_1D_2, \triangle QE_1E_2, \triangle RF_1F_2$ are all equilateral triangles, so:
$$P + \omega D_2 + \omega^2 D_1 = 0 \quad \Rightarrow \quad P + \omega \frac{C+D}{2} + \omega^2 \frac{B+D}{2} = 0 \quad (1)$$
$$Q + \omega E_2 + \omega^2 E_1 = 0 \quad \Rightarrow \quad Q + \omega \frac{A+E}{2} + \omega^2 \frac{C+E}{2} = 0 \quad (2)$$
$$R + \omega F_2 + \omega^2 F_1 = 0 \quad \Rightarrow \quad R + \omega \frac{B+F}{2} + \omega^2 \frac{A+F}{2} = 0 \quad (3)$$
Taking $(1) + (2) \times \omega + (3) \times \omega^2$, we get:
$$(P + \omega Q + \omega^2 R) + \left( \frac{B+C}{2} + \frac{C+A}{2}\omega + \frac{A+B}{2}\omega^2 \right) = \frac{1}{2}(D + \omega E + \omega^2 F) = 0$$
Hence:
$$\begin{aligned} & M_1 + \omega M_2 + \omega^2 M_3 \\ &= \frac{1}{3}\left\{(P + \omega Q + \omega^2 R) + \left(\frac{B+C}{2} + \frac{C+A}{2}\omega + \frac{A+B}{2}\omega^2\right) + D + \omega E + \omega^2 F\right\} \\ &= 0, \end{aligned}$$
Therefore $\triangle M_1M_2M_3$ is an equilateral triangle.