Please change the topic of the thread. It takes away the fun of the problem.

Absolutely Trivial We will use directed angles $\measuredangle$ modulo $\pi$ . By Pascal on $CABFFD$ it suffices to show $C,D,E$ collinear.But $-\measuredangle ADE=\measuredangle ADB=\measuredangle ACB=\measuredangle CBA=\measuredangle CDA=-\measuredangle ADC \implies E,D,C$ collinear so we are done.$\blacksquare$

Pascal theorem on $ACDFFB$. then angle chase to get $C,D,E$ collinear.. I got sniped.

Attachments:

$CLAIM:$ $D,E,C$ are collinear $PROOF:$ SInce E is the reflection of B wrt $\overline{AD}$ $\Longrightarrow \angle EDA = \angle BDA $ $But,$ $\overline{AB}=\overline{AC}$ and $A,D,C,B $ concyclic $\Longrightarrow \angle EDA =\angle BDA =\angle BCA = \angle CBA$ $\Longrightarrow \angle CDA + \angle EDA =\angle (180^\circ - \angle CBA) + \angle CBA$ $=180^\circ$ $\therefore E,D,C $ are collinear. Now,By Pascal's theorem on hexagon $FFDCAB,$ $\overline{FF}\cap \overline{AC},\overline{FD}\cap\overline{AB},\overline{DC}\cap\overline{BF}$ are collinear. $\Longrightarrow K,L,E $ are collinear. [A.W.D]

Attachments:
turkey tst 2019 p3.pdf (13kb)

Meh, We will first show ,$C,D,E$ collinear. $EDA=BDA=ACB=CBA=180-ADC$ Now, just pascal on $ABFFDC$ finishes.

Directed angles mod 180 and pascal's on hexagon CDFFAB unfortunately kills this problem in 1 line. Welp.