Problem 1586

1. Applying Miquel's Theorem:
According to Miquel's Theorem, the circumcircles of triangles $AB'C'$, $BC'H$, and $B'CH$ intersect at a common point $I$. This establishes the existence of the point $I$ where these three circumcircles intersect.

2. Defining Midpoints and Points:
Let $M$ be the midpoint of the segment $\overline{B'C'}$. We need to show that the line $HI$ passes through $M$ and that $B'CHI$ is cyclic.

3. Intersection of Line and Circumcircle:
Let $I' = \overline{HM} \cap (ABC)$. We aim to prove that $B'CHI'$ is cyclic.

4. Symmedian Property:
Note that $\overline{AI'}$ is a symmedian in $\triangle AB'C'$. This can be shown by reflecting over the perpendicular bisector of $\overline{B'C'}$ or by using $\sqrt{bc}$ inversion and angle chasing.

5. Similarity of Triangles:
We have $\triangle IB'M \sim \triangle I'AC'$. This similarity implies that the angles are preserved, specifically:
$$\measuredangle HI'B' = \measuredangle MI'B' = \measuredangle C'I'A = \measuredangle C'B'A = \measuredangle HCB'$$
This chain of equal angles confirms that $B'CHI'$ is cyclic.

6. Conclusion:
Since $B'CHI'$ is cyclic, and $I'$ lies on the circumcircle of $\triangle ABC$, it follows that $I = I'$. Therefore, the line $HI$ passes through the midpoint $M$ of the segment $\overline{B'C'}$.

$$\blacksquare$$