Prove that lines $AM, EF$ and $DI$ meet on $K$ Given a triangle $\triangle ABC$ whose incenter is $I$, its contact triangle is $\triangle EDF$ as in image below and $M$ is the midpoint of $BC$. Then the lines $AM, EF$ and $DI$ meet on point $K$

This problem doesn't seem hard, it may be a simple consequence of $I$ being the circumcircle of $\triangle DEF$ but I'm struggling a bit more than I should.
I thought about finding the polar line of point $K$ wrt to the incircle and it seems to be a  parallel to $BC$ through $A$, then the problem could be transformed in "the polar of $M$ meets $EF$ on the polar of $K$".
Also, does anyone know the name of this result?
This link might help: https://artofproblemsolving.com/community/c776104h1954097
A plagyogonal system could probably solve it fast.
 A: Not sure about the name, but there is a cute synthetic approach.
Let $L$ be the second intersection of $DI$ and the incircle and consider the tangent to the incircle that passes through $L$ - call it $l$, as well as the intersections of $l$ with $AB$ and $AC$ - call them $G \in AB$ and $H\in AC$
Now $l \perp DI$ so $l$ is parallel to $BC$ - meaning $BGHC$ is a trapezium. In particular, the intersection $K'$ of its diagonals lies on the bimedian connecting the midpoint of $BC$ and the midpoint of $GH$ - but this line is precisely $AM$
Further, by construction, the quadrilateral $BGHC$ has an incircle  touching the sides at $D,E,L$ and $F$ - as a consequence of Brianchon, the diagonals $BH$, $GC$ and the lines $DL$ and $EF$ all meet at a point - but since $AM$ passes through $BH\cap GC$ this yields our desired concurrency
Note on Brianchon: It says whenever a hexagon ABCDEF is circumscribed to a conic, the diagonals $AD$, $BE$ and $CF$ meet at a point. It is actually the dual of Pascal's (maybe more well known) Theorem that is concerned with and inscribed hexagon $ABCDEF$ and states that the $3$ intersection points $AB\cap DE , BC\cap EF$ and $CD\cap AF$ are colinear
In our case, we're actually applying Brianchon twice to deduce the four desired lines meet, and we're actually applying it in a degenerate case(holds by continuity) : once to $BDCGLH$ and once to $BEGHFC$
Hope this helps
