Triangle and its circumcircle Let $ABC$ be a triangle and $\Gamma$ its circumcircle. On sides $AC$, $BC$ lies respectively points $E$, $F$ such that $CE=BE$ and $CF=AF$. $CM$ is a median of triangle $EFC$. Show that line $CM$ pass throught point $K$ which is point where meets tangents to circle $\Gamma$ at points $A, B$. 
Here is my sketch, I know that $E, F, A, B$ lies on a common circle because triangle $EFC$ and $ABC$ are similar, but don't know what to do now.

 A: Because $\triangle AFC$ and $\triangle BEC$ are isosceles, we have
$$\angle C \cong \angle CAF \cong CBE$$

(Note that $\angle C$ is necessarily acute.) Moreover, the bisectors of $\angle AEB$ and $\angle BFA$ create four more copies of $\angle C$. That two of these copies are $\angle HEA$ and $\angle HFB$ implies that $\square HEFC$ is a parallelogram, the diagonals of which bisect each other. Thus, $\overline{CH}$ passes through $M$.

It remains to show that $H$ coincides with $K$.
Note that $\angle EHF$ is yet another copy of $\angle C$ (as the opposite angle in the parallelogram). Since $\angle EAF$ and $\angle EBF$ are too, the points $H$, $A$, $B$ are concyclic with $E$ and $F$ (because they subtend the same angle with chord $\overline{EF}$).

This implies that $\angle ABH$ (which must be congruent to $\angle AFH$, as both subtend chord $\overline{AH}$) is also a copy of $\angle C$; likewise, $\angle BAH \cong \angle C$.

We see, then, that $H$ is the apex of the[*] isosceles triangle with base segment $\overline{AB}$ and base angles congruent to $\angle C$. This description also fits $K$ (why?), so the points must coincide.


[*] There are, of course, two such isosceles triangles with base $\overline{AB}$. "The" triangle in question is the one on the opposite side of $\overleftrightarrow{AB}$ from $C$. (Note that $\angle C$'s acuteness is key to making this distinction.)
A: (These are complete steps, you just need to fill in the minor details.)
Step 1: Show that $AEFBK$ is con cyclic, as drawn in your diagram. It could be easier to show that $O$, the center of $\Gamma$, also lies on this circle.
Step 2: Show that $EF=KB=KA$, because they each subtend the same angle in this circle.
Step 3: Show that $KE \parallel BF$ and $KF \parallel EA$.
Step 4: Hence show that $KECF$ is a parallelogram.
Hence $CK$ bisects $EF$, so $CKM$ is a straight line.
