Let $AH_1$, $BH_2$, $CH_3$ be the altitudes of $\triangle ABC$ (Fig. 29), $M$ is the midpoint of $BC$, and let $O_1$ be the circumcenter of $\triangle YHX$. We want to show that $\triangle YHX \sim \triangle ABC$.

Since $\angle BH_3C = \angle YDC = 90^\circ$, $\angle ABC = \angle HYX$. Similarly, $\angle ACB = \angle HXY$. Therefore, $\triangle YHX \sim \triangle ABC$.

Let $HH'$ be the altitude of $\triangle YHX$, and let $M'$ be the midpoint of $XY$.

Clearly, $O_1M' \perp XY$, $OM \perp BC$. Thus, $HH'$ and $AH_1$ are the corresponding elements in similar triangles. Similarly, $O_1M'$ and $OM$ are the corresponding elements in similar triangles. Thus,
$$\frac{HH'}{AH_1} = \frac{O_1M'}{OM}.$$
From the other side, it is clear that $\triangle OMD \sim \triangle AH_1D$. Hence,
$$\frac{H_1D}{AH_1} = \frac{MD}{OM}.$$
These two equalities, together with $HH' = H_1D$, since $HH_1DH'$ is a rectangle, give $O_1M' = MD$. Thus, $O_1M'DM$ is a rectangle. Therefore, $O_1M \perp BC$, so $O_1$ belongs to the perpendicular bisector of $BC$. Hence, $O_1B = O_1C$.