, 2022

Lemma. Let $ABC$ be a triangle whose circumcenter is $O$. Let $\ell$ be an arbitrary line passing through $O$. Let $Y, Z$ be two points on $\ell$ such that $AY, AZ$ are perpendicular to $AC, AB$, respectively. Let $P, Q$ be the projections of $Y, Z$ onto $AB, AC$, respectively. Then $PQ$ bisects $AB$.

Proof of Lemma. Let $M_a, M_b, M_c$ be midpoints of sides $BC, CA, AB$, respectively. Then we can see that (using signed segments),
$$\frac{PM_c}{M_cA} = \frac{YO}{OZ} = \frac{AM_b}{M_bQ}$$
because $PY$, $M_cO$, $AZ$ are parallel and $AY$, $M_bO$, $QZ$ are parallel. Therefore
$$\frac{BP}{PA} = \frac{M_cP - M_cB}{PM_c + M_cA} = \frac{M_cA - PM_c}{PM_c + M_cA} = \frac{AM_b - M_bQ}{AM_b + M_bQ} = -\frac{CQ}{QA}.$$
By Menelaus' theorem, we thus know that $PQ$ bisects $AB$. $\Box$

Alternative proof of Lemma. We still have $PM_c/M_cA = AM_b/M_bQ$. Let $PM_a$ intersects with $AC$ at $Q'$, and let $QM_a$ intersects with $AB$ at $P'$. Since $M_aM_c$ is parallel to $AC$, we have $PM_c/M_cA = PM_a/M_aQ'$. Similarly $AM_b/M_bQ = P'M_a/M_aQ$. This then shows that $PM_aQ'$ and $P'M_aQ$ are indeed the same line, as desired. $\Box$

To prove the original statement, suppose that $AH$ intersects with $BC$ at $D$. It is well-known that $HX = 2HD$. Let $M_a$ be the midpoint of $BC$, then we have $HX = 2HD = 2OM_a = AH$, showing that $D$ is the midpoint of $HX$, and $H$ is the midpoint of $AX$. Since $OH$ is parallel to $BC$ and $HX = 2DX$, we know that $YX = 2BX$ and $ZX = 2CX$. Thus $AY \parallel BH \perp AC$ and $AZ \parallel CH \perp AB$. Now take $\ell$ to be $OH$ and apply the Lemma. We then know that $PQ$ bisects $BC$, as desired. $\Box$