Solution:

Let $\alpha = \angle BAC$. Let $BB'$ and $CC'$ be altitudes in triangle $ABC$, as shown.

Claim. Triangle $CDE$ is isosceles with $CE = DE$.

Proof.

$$\angle DEC = \angle DOC \\ \qquad = \angle BOC \\ \qquad = 2\angle BAC \\ \qquad = 2\alpha$$

$$\angle DCE = \angle C'CA = 90^\circ - \angle C'AC \\ \qquad = 90^\circ - \alpha$$

$$\begin{array}{rl} & \angle CDE = 180^\circ - \angle DEC - \angle DCE \\ & \qquad = 180^\circ - 2\alpha - (90^\circ - \alpha) \\ & \qquad = 90^\circ - \alpha \\ & \qquad = \angle DCE. \end{array} \quad \text{(angle sum in } \triangle CDE)$$

Since $CE = DE$ and $CE = FE$, we have $CE = DE = FE$, therefore $E$ is the circumcentre of triangle $CDF$. Consequently,

$$\begin{array}{rl} & \angle HDF = 180^\circ - \angle CDF \\ & \qquad = 180^\circ - \frac{1}{2} \cdot \angle CEF \\ & \qquad = 180^\circ - \frac{1}{2} \bullet (180^\circ - 2\angle ECF) \\ & \qquad = 180^\circ - \frac{1}{2} \bullet (180^\circ - 2\angle B'CB) \\ & \qquad = 180^\circ - (90^\circ - \angle B'CB) \\ & \qquad = 180^\circ - \angle B'BC \\ & \qquad = 180^\circ - \angle HBF \end{array} \quad \text{(angles in } \triangle B'BC)$$

so $B H D F$ is cyclic.