Problem 1443

1. Define Points and Midpoints:
- Let $S$ be the midpoint of $BC$.
- Let $K$ be the intersection of $BH$ and $DE$.

2. Angle Chasing:
- We need to show that $B, D, K, F$ lie on a circle. To do this, we will use angle chasing and properties of reflections and midpoints.
- Since $D$ and $E$ are midpoints of $AB$ and $AC$ respectively, $DE$ is the midline of $\triangle ABC$, parallel to $BC$ and half its length.

3. Reflection Properties:
- $F$ is the reflection of $H$ with respect to $DE$. This means $DE$ is the perpendicular bisector of $HF$, and thus $DH = DF$ and $EH = EF$.

4. Angle Calculation:
- We need to show that the angles $\angle DBF$ and $\angle DBC$ are equal modulo $\pi$.
- Consider the angles:
$$(DB; DF) \equiv (DB; DE) + (DE; DF) \equiv (BA; BC) + (DH; DE)$$
- Since $DE \parallel BC$, we have:
$$(DH; DE) \equiv (DH; DS) + (DS; DE) \equiv (SD; SE) + (CA; CB)$$
- Using the fact that $S$ is the midpoint of $BC$, we get:
$$(SD; SE) \equiv (AC; AB)$$
- Combining these, we have:
$$(DB; DF) \equiv (BA; BC) + (AC; AB) + (CA; CB) \equiv 2(CA; CB) \equiv 2(EA; ED)$$
- Since $F$ is the reflection of $H$ across $DE$, we have:
$$(EA; EF) \equiv (KB; KF) \mod \pi$$

5. Cyclic Quadrilateral:
- From the above, we conclude that $B, D, K, F$ lie on a circle because:
$$(DB; DF) \equiv (KB; KF) \mod \pi$$

6. Isogonal Conjugates:
- Since $B, D, K, F$ lie on a circle, we have:
$$(BF; BD) \equiv (KF; KE) \equiv (KE; KH) \equiv (BC; BH) \mod \pi$$
- This implies that $BF$ and $BH$ are isogonal conjugates with respect to $\triangle ABC$.

7. Conclusion:
- Since $BF$ and $BH$ are isogonal conjugates, $BF$ passes through the circumcenter of $\triangle ABC$.

$\blacksquare$