Note that there is a circle with center $E$ that is tangent to $KL$ and tangent to $BA,BC$ at $A,C$. Let $EL$ intersect $AC$ at $N'$. By angle chasing, we get $\angle KAN'=\angle KEN'$, so $KAEN'$ is cyclic, so $\angle KN'E=\angle KAE=90$. So $MK=ML=MN'$, and this implies that $MN'\parallel BC$. So $N\equiv N'$ and we have $\angle KNL=90$.

let the parallel through $K$ to $BC$ intersect $AC$ at $S$ than $AK/KS=AB/BC=1$ so in trapezoid $KSCL$ $MN=(KS+LC)/2=(AK+LC)/2=KL/2=MK=ML$ so $\angle KNL=90$.

Very cute problem. Let $B'$ be the reflection of $B$ over the perpendicular line to $\overline{BC}$ through $K$. Hence $$MN = \frac 12(KB'+LC) = \frac 12(KB+LC) = KM = ML,$$and thus $\angle KNL = 90^\circ$.