Let $O_1,O_2$ the incenters , $D$ be the tangency point of the incircle of $ABC$ with $AC$ we have $\frac{AD}{AC}=\frac{AB+AC-BC}{AC}$, $\frac{AH_1}{AB}=\frac{AB+AH-BH}{AB}=\frac{AC\cos\angle A+AB \cos \angle A-BC\cos \angle A}{AC\cos\angle A}=\frac{AD}{AC}$ hence $DH_1 \parallel BC $ idem $DH_2\parallel AB$ thus $BH_1H_2D$ is rectangle and $O$ is the midpoint of $BD$ but $BHD$ right-angled triangle then $OD=OH(^*)$. More by the same trick we deduce $B_1H=DB_2$ which implies $O_1O_2,HD$ has the same bisector so from $(^*)$ we get $O$ is on the bisector of $B_1B_2$ RH HAS