$AH$ 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 $I$ and $IM \perp AB, IN \perp AC$ $(M \in AB, N \in AC)$.
We have that $(BF, BM) = (CF, CN) = (DF, DA)$ and $\dfrac{\overline{BF}}{\overline{BM}} = \dfrac{\overline{CF}}{\overline{CN}}$.
(because $FE$ being the bisector of $\angle CFB \implies \dfrac{\overline{CF}}{\overline{FB}} = \dfrac{\overline{CE}}{\overline{EB}}$ alongside $\dfrac{\overline{CE}}{\overline{CN}} = \dfrac{\overline{BF}}{\overline{BM}} = 1$).
$\implies \triangle FBM \sim \triangle FCN \implies (MF, MA) = (NF, NA) \implies M, F, N, A$ are concyclic.
Furthermore, we have that $M, N \in (AI) \implies F \in (AI) \implies FA \perp FI$.
It is evident that $\overline{KF} \cdot \overline{KA} = \overline{KB} \cdot \overline{KC} \implies K$ lies on the radical axis of $(AI)$ and $(CIB)$.
$\implies KI \perp AD$. However, $KH \perp AH \implies A, I, H, K$ are concyclic.
For now, we work backwards.
Let $FF' \parallel BC, F \in (ABC)$, we have that $F(JKMF') = -1 \impliedby F(LAFF') = -1$
But that's all.
Best Answer
What about this solution (instead of searching for a credible and/or official solution that you may never find to give it the bonus reputation)?
Like the capitals you've associated to points, let $I$ be the center of inscribed circle inside $\bigtriangleup ABC$. Take $S$ the intersection of $AD$ and $BC$, then we have:
$$\dfrac {AI} {IS} = \dfrac {\sin \angle DAB}{\sin \angle CSA}= \dfrac {\sin \angle SCD}{\sin \angle CSD}= \dfrac{DC}{DS} =\dfrac {DI} {DS}$$
Let $P$ be the intersection of line $\overline {IE}$ and $DL$ and $Q$ be the intersection of $IH$ and $DF$. By Seva's theorem in $\bigtriangleup IDH$ and $IP \| AH $ and above equation we have:
$$\dfrac {IQ}{QH}= \dfrac {PD}{HP} . \dfrac {IS}{DS}= \dfrac {DI}{AI} . \dfrac {AI}{DI}=1$$
So $Q$ is the midpoint of $IH$. let $R$ the intersection of $AH$ and $DF$, then $IEHR$ is a rectangle. From the fact that the quadrilateral $ASEF$ is cyclic (why?) and $RI \| BC$, we conclude $AIRF$ is cyclic, so as you mentioned $IF \bot AK$, therefore $IEKF$ and then $AIHK$ is cyclic and we have $KI\bot AD$ and $\angle HID = \angle AKJ$.
Ultimately, $\bigtriangleup HID \sim \bigtriangleup JKF$ and the tangent of $(ABC)$ at $F$ bisects $KJ$ like the way $DF$ bisects $IH$ at $Q$.
I think other solutions may use the cyclic characteristics of quadrilaterals $ASEF$ and $FEHL$.