Prove that the tangent of $(ABC)$ at point $F$ passes through the midpoint of $KJ$.

$AH$ is a altitude of $\triangle ABC$, $AD$ is the bisector of $\angle CAB$ and $BC$ meets the incircle of $\triangle ABC$ at point $E$ and $(D \in (ABC), H \in BC)$. Let the intersections between $DE$, $DH$ and $(ABC)$ are respectively $F$, $L$ $(L \not\equiv D \not\equiv F)$. $AF$ and $FL$ cut $BC$ correspondingly at $K$ and $J$. Prove that the tangent of $(ABC)$ at point $F$ passes through the midpoint of $KJ$.

Let the incenter of $\triangle ABC$ be $I$.

We would have that $IF \perp AK$ and $H \in (AIK)$

$\implies \angle KFI = \angle KIA = 90^\circ \implies \angle FAD = \angle FIK$

But that's all.

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