First, since $CD \perp AB$ and $DK$ and $DN$ are angle bisectors to the right angles
$\angle \, ADC$ and $\angle \, BDC$, then $$\angle \, KDN = \angle \, KDC + \angle \, NDC = 45^{\circ} + 45^{\circ} = 90^{\circ}$$
However, $\angle \, KCN = 90^{\circ}$ so the quadrilateral $CKDN$ is inscribed in a circle.
Next, prove that $KL\, || \, CB$ and $NL\, || \, CA$ using the properties of angle bisectors and the similarity between triangles $ABC, ACD$ and $BCD$. Indeed, since $DK$ is a bisector of the angle at vertex $D$ of triangle $\Delta \, ADC$, we apply the theorem that
$$\frac{AK}{KC} = \frac{AD}{DC}$$ But triangles $\Delta \, ACD$ is similar to $\Delta \, ABC$ so $$\frac{AD}{DC} = \frac{AC}{CB}$$ so
$$\frac{AK}{KC} = \frac{AC}{CB}$$ By the fact that $CL$ is an angle bisector of the angle at vertex $C$ of triangle $\Delta\, ABC$
we have that
$$\frac{AC}{CB} = \frac{AL}{LB}$$ so consequently
$$\frac{AK}{KC} = \frac{AL}{LB}$$
which by Thales' intercept theorem implies that $KL \, || \, CB$. Analogously, one can show that $NL \, || \, CA$.
Then quad $CKLN$ is a rectangle, so $KL =NL$ as diagonals in a rectangle. Therefore the point $L$ also lies on the circumcircle of quad $CKDN$ and $KN$ and $CL$ are diameters of the said circle.
Observe a triangle $ACL$ and extend halfline $AC$ to $X$.
Notice that $CT$ is outer angle bisector for angle $\angle ACL$ since $\angle LCT = \angle LCX = 60^{\circ}$.
Since $T$ is also on angle bisector $AT$ of angle $\angle CAL$. So $T$ is at equal distance to lines $CL$ and $LB$, so $LT$ is angle bisector of angle $\angle CLB$.
Best Answer
Let $AC = 2x$. Since triangle $AHC$ is half of the equliateral triangle with side $2x$ (reflect $C$ across $AB$) we see that $CH =x$.
Let $y= AB/6$ then $AH =3y$ and $AM = 2y$ so $MH = y$ and thus $${AC\over CH} = {AM\over MH}$$
so by angle bisector theorem $CM$ is angle bisector for $\angle ACH$.