Let $ O $ be the circumcenter of $ \triangle ABC $. It's well-known that $ AH=2 \cdot OM $. Notice that $ O_2 $ is the reflection of $ O $ respects to $ BC $ since the reflection of $ H $ respects to $ BC $ lies on the circumcircle of $ \triangle ABC $ Thus, we have $ O_2M=OM=\frac{1}{2} \cdot AH=AO_1 $, then it's trivial that the conclusion follows.

Working over $\mathbb{C}$, it suffices to show that $o_1+m = o_2+a$. Placing the circumcenter of $ABC$ at the origin, we may take \[o_1=\frac 12\cdot (h+a) = a+\frac{b+c}{2}.\] Since $\angle BHC = \pi - \angle A$, the point $O_2$ is the just the reflection of the circumcenter of $ABC$ in line $BC$, which is the same as the the reflection of the origin in $m$. Thus, $o_2 = b+c$. Then \[o_1+m = a+\frac{b+c}{2}+\frac{b+c}{2} = a+b+c = (b+c) + a = o_2+a\] as desired$.\;\blacksquare$

Dear Mathlinkers, posted before http://www.artofproblemsolving.com/community/c6t48f6h1084452_aze_jbmo_tst Sincerely Jean-Louis

Easy complex bash $o_1+m=^{?}o_2+a$ $o_1=\frac{a+h}{2}=a+\frac{b+c}{2}$ $m=\frac{b+c}{2}$ By lemma $o_2=b+c$ $o_1+m=a+b+c$ $o_2+a=b+c+a$ $\to o_1+m=o_2+a$ qed.