I write my solution briefly, By Euler theorem we can find the circumcenter of $(ABC)$ called $O$. We can also find the excenter of vertex $A$ called $I_A$. (note that $BI$ is bisector of $\angle HBO$ therefor $(OH,II_A)=-1$.) Let $T$ be the midpoint of $II_A$. Points $B,C$ are the intersections of circle with center $O$, radius $OT$ and circle with centere $T$, radius $IT$. And $A$ is the intersection of line $OH$ and circle with center $O$,radius $OT$.