# 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 $$I$$ 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.

• you just need to add my last statement to your findings and then the proof is done: $\angle HID = \angle AKJ$. Ultimateley $\bigtriangleup HID \sim \bigtriangleup JKF$ and the tangent of $(ABC)$ at $F$ bisects $KJ$ like the way $DF$ bisects $IH$ at $Q$.
– MasM
Commented Oct 4, 2019 at 0:40

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 $$FEHL$$.