Winter Mathematical Competition

We may assume that $AC<BC$. Denote by $k$ the circumcircle of $△KCL$ and by $K_1$ and $L_1$ the midpoints of $AC$ and $BC$, respectively. Then $K{K}_1\parallel L{L}_1$, i.e. $K{K}_1{L}_{1}L$ is a trapezoid. If $T$ is the midpoint of $CM$ then $$CM=\frac{BC-AC}{2},CT=\frac{BC-AC}{4},$$ $$K_{1}T=\frac{AC}{2}+\frac{BC-AC}{4}=\frac{BC+AC}{4},$$ $$L_{1}T=\frac{BC}{2}-\frac{BC-AC}{4}=\frac{BC+AC}{4}.$$ Therefore $T$ is the midpoint of $K_1{L}_1$.

Let $t$ be the segment whose ends are the midpoints of $KL$ and $K_1{L}_1$. Then $T\in t$, $t\perp K_1{L}_1$ 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. $KA\perp LB$.