Consider $AD$, $BE$ and $CF$ being altitudes of $\triangle ABC$ where $D \in BC$, $E \in CA$ and $F \in (A, B, C)$. $FD$ extended intersects $(A, B, C)$ at $K$. Prove that $AK$ passes through the midpoint of $DE$.
Here's what I've done.
Let the intersections of the tangent of $(A, B, C)$ at $C$ and $AK$, $AD$ respectively $G$ and $H$.
We have that $$\widehat{GCK} = \widehat{GAC} \implies \triangle GCK \sim \triangle GAC \implies \dfrac{GC}{GA} = \dfrac{GK}{GC} \implies GK \cdot GA = GC^2$$
Now I just need to prove that $GD^2 = GK \cdot GA$ so that $GC = GD \implies \widehat{GCD} = \widehat{GDC}$
$\implies \widehat{GHD} = \widehat{GDH} \implies GH = GD \implies GC = GH$.
Then I will let the intersection of $AK$ and $DE$ be $I$.
We have that $AEDB$ is a cyclic quadrilateral because $\widehat{ADB} = \widehat{AEB} = 90^\circ$.
$\implies \widehat{BAE} + \widehat{BDE} = 180^\circ \implies \widehat{BAC} = \widehat{CDE}$.
And $\widehat{BAC} = \widehat{BCH}$ because $CH$ is a tangent of $(A, B, C)$ at $C$.
$\implies \widehat{CDE} = \widehat{BCH} \implies DE \parallel CH$.
Using the intercept theorem for $DI \parallel HG$ and $EI \parallel CG$.
We have that $\dfrac{DI}{HG} = \dfrac{AI}{AG}$ and $\dfrac{EI}{CG} = \dfrac{AI}{AG}$ $\implies \dfrac{DI}{HG} = \dfrac{EI}{CG}$.
But $HG = CG \implies DI = EI$.