Denote by $D$ midpoint of $AC$. We have that $MO \mid \mid BH$. From yhis is $\angle MOG= \angle GHB$. It is also $\angle BGH= \angle DGO \leq MCO = \angle MOG$. Hence $BH\leq BG$. Equality holds if $\angle DGM=0$, but then points $B,H,G$ lies on same line. SO there must be $H = G \Longleftrightarrow AB=AC=BC$. q.e.d.

Ok, now realized my solution has some flaws in it.

what is tagiyev's pisi lemma?can you send me this lemma?

another proof: Call O the circumcenter, K the projection of B on OH and $ \theta$ the angle between the euler line and AC. So we have $ \sin \theta = \frac{\overline{GO}}{2r}$ then $ \overline{KH} = \overline{BH} \cdot \sin \theta = 2R \cos \beta \sin \theta = \overline{GO}\cos \beta \le \overline{GO} = \frac{\overline{GH}}{2}$. Then $ \overline{KH} \ge \overline{KG} \Longleftrightarrow \angle GHB \ge \angle HGB \Longleftrightarrow \overline{BG} \ge \overline{BH}$

Let $ K$ be the midpoint of segment $ AC$ and $ O$ be the circumcenter of triangle $ ABC$.$ BGH$ similar to $ KGO$ and $ BG\geq BH$ iff $ GK\geq OK$. $ MG=MO=R$ and by the triangle inequality $ GK+MK\geq MG$ , $ GK+R-OK\geq R$ and $ GK\geq OK$ . The proof is completed.