In the solution of this problem, we will be using the following well-known fact: in any triangle, the distance from the vertex to the orthocenter is twice larger than the distance from the circumcenter to the opposite side.

Denote by $O_B$ and $H_B$ the projections of $O$ and $H$ correspondingly onto the line $AC$, define points $O_C$ and $H_C$ similarly (fig. 9). From the fact above it follows that $BH = 2OO_B$ and $CH = 2OO_C$. Also note, that $OO_B \parallel HH_B$ and $HO = OY$, and therefore $OO_B$ is the midline of $\triangle YHH_B$, so $HH_B = 2OO_B = BH$. Similarly $HH_C$ is the midline of $\triangle XOO_C$, so $HH_C = \frac{1}{2}OO_C = \frac{1}{4}CH$. As the quadrilateral $BHC_HC_B$ is cyclic, $BH \cdot HH_B = CH \cdot HH_C$, and $BH^2 = 4HH_C^2$, so $BH = 2HH_C$, and $\angle HBH_C = 30^\circ$. Now it's clear that $\angle BAC = 60^\circ$.