$(A, AD)$ intesects $(ABC)$ at $E$ and $F$ where $AD$ is the altitude of $\triangle ABC$. Let $D'H \parallel EF$ and $D'$ is the reflection of $D$ in $G$, which is the foot of the perpendicular line to $EF$ from $D$ $(H \in BC)$. $I$ is the reflection of $D$ in point $A$. Let $K$ be the midpoint of $DJ \perp HI$ $(J \in HI)$. Prove that $K \in (ABC)$.
Because $A$ is the midpoint of $DI$, we might need to prove that $AK \parallel IJ \implies AK \perp DK$
$ \implies K, D$ and $L$ are collinear where $AL$ is the diameter of $(ABC)$.
Further than that, I don't know how.