Any solution ?

bump

Wow complex numbers actually work Set $(ABC)$ as the unit circle and denote point $X$ as the complex number $x$ Since $MP \perp ON$ we can write $$\frac{m-p}{n}=\frac{\overline{m}-\overline{p}}{\overline{n}}$$And since $O-P-N$ are collinear $\frac{m}{n}=\frac{\overline{m}}{\overline{n}}$ which gives us $$\frac{p}{n}=\frac{\frac{m}n+\frac{\overline{m}}{\overline{n}}}{2}$$Now $M'PMD$ is a parallelogram, so $$m'+h=\frac{b+c}2+p$$which is the same as $$m'=\frac{a+b+c}{2}-\frac{1}{2} \cdot \frac{m+a}{a\overline{m}+1}$$after some simplification Finally notice that $$\mid \frac{m+a}{a\overline{m}+1} \mid = 1$$since it's conjugate multiplies to $1$ which means that $\mid m'-n_9 \mid=\frac{1}{2}$ which means $m' \in (DEF)$ as desired $\blacksquare$