Proving that given a triangle $ABC$ the line $AB$ is parallel to $CK$ Supposing a triangle $ABC$, where $|AC|\neq |BC|$, denote the incentre $I$ and the points of tangency between the inscribed circle and $BC$,$CA$,$AB$ to be $D,E,F$ respectively. $M$ is the midpoint of $AB$. $K$ is the intersection of the perpendicular to $CM$ passing through $I$, and line $DE$. The task is to prove that $CK||AB$.
I have realized that $DE$ is a side of the Gergonne triangle of $ABC$, but I haven't been able to connect it with angle bisectors or anything pertaining to the incentre, nor with medians. I have also tried constructing the circumscribed circle and looking at the perpendicular bisectors but without much success.
I'd really appreciate your help.
 A: It's well known that $DE, IF, CM$ are concurrent. Call this point $X$.
Then $CX\perp IK$, $CI\perp XK$, meaning $I$ is the orthocenter of $CXK$ and the result follows.
Another interesting thing is that $\triangle CXK$ is self-polar wrt the incircle.
Edit:

The following proof that $DE,IF,CM$ are concurrent is not mine. It's taken from Evan Chen's EGMO.
Let $X$ be the intersection of $DE$ and $FI$ and let $A',B'$ be points on $CA$ and $CB$ such that $A'B'\parallel AB$ and $A'B'$ passes through $X$.
Now $E$ is the foot of $I$ onto $CA'$, $X$ is the foot of $I$ onto $A'B'$ ($B'A'\parallel BA$ and $IF\perp AB$, so $IF\perp AB$), and $D$ is the foot of $I$ onto $CB'$. Then $EXD$ is in fact the Simson line of $I$ wrt $\triangle CA'B'$, which means that $I$ is on the circumcircle of $\triangle CA'B'$.
Since $CI$ is an angle bisector of $\angle A'CB'$, $I$ is the arc midpoint of arc $B'A'$, and so $XI$ is the perpendicular bisector of $B'A'$. This means $X$ is the midpoint of $B'A'$, and so taking a homothety at $A$ which takes $A'B'\mapsto AB$ shows that the $C$-median passes through $X$ as well.
Once seeing $I$ is the orthocenter of $\triangle CXK$, $CK\perp IX$ but we already know $IX\perp AB$ so $CK\parallel AB$.
