Construct the circumcircle of $\triangle ABC$. Let $D'$, $E'$ and $F'$ denote the points where the altitudes meet the circumcircle. Let $K$ be the intersection of $AD'$ with $BC$ and denote the orthocentre of $\triangle ABC$ by $H$. Then $\angle BAD' = 90^\circ - \angle ABC = \angle BCF$ and $\angle BAD' = \angle BCD'$, hence the triangles $HKC$ and $KD'C$ are congruent. This shows that $|HK| = |KD'|$ and $D' = D$ is the image of $H$ when reflected in $BC$. Similarly, $E' = E$ and $F' = F$.

Now $\angle DFC = \angle DAC = 90^\circ - \angle ACB$ and $\angle CFE = \angle CBE = 90^\circ - \angle ACB$. Therefore, $\angle DFC = \angle CFE$ which means that $CF$ is the bisector of $\angle DFE$. Similarly, $BE$ and $AD$ are the bisectors of $\angle FED$ and $\angle FDE$, respectively. This shows that $H$ is the incentre of $\triangle DEF$.