We may assume that $AC < BC$. Denote by $k$ the circumcircle of $\triangle KCL$ and by $K_1$ and $L_1$ the midpoints of $AC$ and $BC$, respectively. Then $KK_1 \parallel LL_1$, i.e. $KK_1L_1L$ is a trapezoid. If $T$ is the midpoint of $CM$ then
$$CM = \frac{BC - AC}{2}, \quad CT = \frac{BC - AC}{4},$$
$$K_1T = \frac{AC}{2} + \frac{BC - AC}{4} = \frac{BC + AC}{4},$$
$$L_1T = \frac{BC}{2} - \frac{BC - AC}{4} = \frac{BC + AC}{4}.$$
Therefore $T$ is the midpoint of $K_1L_1$.

Let $t$ be the segment whose ends are the midpoints of $KL$ and $K_1L_1$. Then $T \in t$, $t \perp K_1L_1$ and $t$ is a diameter of $k$, since $T$ is the midpoint of the chord $CM$ and $t \perp CM$.
If $t \cap KL = O$, then $O$ is the midpoint of the chord $KL$. Hence we have two possibilities:
1) $t \perp KL$ – then $KL \parallel AB$;
2) $KL$ is a diameter of $k$. Then $\angle KCL = 90^\circ$ and $\angle KCA + \angle LCB = 90^\circ$. But $\angle KAC = \angle KCA$ and $\angle LBC = \angle LCB$, which implies that $\angle KAC + \angle LBC = 90^\circ$, i.e. $KA \perp LB$.