China Hong Kong Mathematical Olympiad

Let $P$ be the midpoint of $DF$. Then $\angle FPB = 90^\circ$. Since $\angle BFP = \angle DEH$, we have $\triangle BPF \sim \triangle DHE$. This implies $\frac{BF}{PF} = \frac{DE}{HE}$, which yields
$$BF \times HE = DE \times PF = \frac{1}{2}DE \times DF.$$
By symmetry,
$$CE \times HF = \frac{1}{2}DE \times DF = BF \times HE.$$
Together with $\angle BFH = 90^\circ + \frac{A}{2} = \angle CEH$, we obtain $\triangle BFH \sim \triangle CEH$. It follows that $\angle FBH = \angle ECH$.

Let $B'$ be the reflection of $B$ in the line $AH$. Note that $B'$ lies on $BC$ since $AH \perp BC$. Also, $B' \neq C$ as $\triangle ABC$ is scalene. Therefore, we find that
$$\angle AB'H = \angle ABH = \angle ACH.$$
This implies $A$, $H$, $B'$, $C$ are concyclic. Thus,
$$\angle HBC = \angle HB'B = \angle HAC = 90^\circ - C,$$
and hence $BH \perp AC$. So $H$ is the orthocentre of $\triangle ABC$.