Problem 1555

1. Define the problem setup and key points:
- Let $ABC$ be an acute, non-isosceles triangle with circumcircle $(O)$.
- $BE$ and $CF$ are the altitudes of $\triangle ABC$, intersecting at $H$.
- $M$ is the midpoint of $AH$.
- $K$ is the point on $EF$ such that $HK \perp EF$.
- A line parallel to $BC$ intersects the minor arcs $AB$ and $AC$ of $(O)$ at $P$ and $Q$, respectively.

2. Identify the tangents and radical axes:
- Let $\ell_F$ and $\ell_E$ be the radical axes of $(ABC)$ with points $F$ and $E$, respectively.
- $PB \cap \ell_F = R$ is on the tangent to $(BPF)$ at $F$.
- $QC \cap \ell_E = S$ is on the tangent to $(CQE)$ at $E$.

3. **Show that $FR \cap ES \in MK$:**
- Identify other points on $\ell_F$ and $\ell_E$:
- If the tangents to $(BFEC)$ at $F$ and $E$ intersect $BC$ at $X$ and $Y$, then $X \in \ell_F$ and $Y \in \ell_E$.
- If the perpendicular bisectors of $BF$ and $CE$ intersect the tangents to $(ABC)$ at $B$ and $C$ at $U$ and $V$, then $U \in \ell_F$ and $V \in \ell_E$.
- By angle chasing, $\angle AFE = \angle ACB \implies U, V \in EF \implies U, F, K, E, V$ are collinear.
- $X, F, M$ are collinear since $XF \perp$ the line through $F$ and the midpoint of $BC$, and $MF$ is also perpendicular to this line ($M$ is the circumcenter of $\triangle AEF$).

4. Use the Ratio Lemma:
- If $R' = FR \cap MK$, then by the Ratio Lemma on $\triangle FMK$ and $\triangle FUX$:
$$\frac{MR'}{R'K} = \frac{FM}{FK} \cdot \frac{XR}{UR} \div \frac{XF}{UF}$$
- Similarly, for $S' = ES \cap MK$:
$$\frac{MS'}{S'K} = \frac{EM}{EK} \cdot \frac{YS}{VS} \div \frac{EY}{EV}$$

5. Relate angles and apply the Ratio Lemma:
- Note that $\angle UBR = \angle VCS$ and $\angle XBR = \angle YCS$.
- By the Ratio Lemma on $\triangle BXU$ and $\triangle CYV$:
$$\frac{XR}{UR} \div \frac{YS}{VS} = \frac{BX}{BU} \div \frac{CY}{CV}$$
- Substituting this in, we get:
$$\frac{FM}{FK} \cdot \frac{BX}{BU} \cdot \frac{UF}{XF} = \frac{EM}{EK} \cdot \frac{CY}{CV} \cdot \frac{VE}{YE}$$

6. Simplify using properties of the triangle:
- Since $UB = UF$, $VC = VE$, and $FM = EM$, this simplifies to:
$$\frac{EK}{FK} = \frac{CY}{EY} \cdot \frac{FX}{BX}$$
- By the Law of Sines on $\triangle CYE$ and $\triangle BXF$, the RHS equals:
$$\frac{\tan \angle C}{\tan \angle B} = \frac{dist(C, AH)}{dist(B, AH)} = \frac{EK}{FK}$$
- This confirms the desired result.

$\blacksquare$