Let $D=BC\cap K, E=AC\cap k$ and $F=AB\cap k$. Also let $G= CI \cap DF$ and $N$ is the midpoint of $AC$. Now by Iran's lemma we get that $MN, CI, DF$ and the circle with diameter $AC$ concur at $G$. Now let $F'$ be the reflection of $F$ across $M$, we get $MF=MT=MF'$ but $GM\|BC$ by the lemma so $MF=MG$ then $G, F, T, F'$ lie on a circle with center at $M$, hence $\angle F'GF= 90^{\circ}$. Now by the lemma $AGLC$ and $AGFIE$ are cyclic so $\angle CGL=\angle CAL$, and $\angle CGD=\angle IGF= \angle IAF=\frac{\angle CAL}{2}$ so $GD$ bisects $\angle CGL$. From here we get that $(F', F; W, L)=-1$ and since $M$ is the midpoint of $FF'$ we get $MF^2=MW\cdot ML=MT^2$ and the conclusion follows.

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Seeing that $(F,F';W,L)=-1$ is much easier with a perspectivity at $C$ to $FI$.

Diagram: https://i.postimg.cc/1RN3Vdbx/image.jpg WLOG, diagram as shown. The desired conclusion is equivalent to $MT$ being tangent to $(WTL)$. By Power of a Point, this is equivalent to $MW \cdot ML = MT^2 = MF^2$ where $F$ is the $C$-intouch point. Since all of $MW$, $ML$, and $MF$ are "known" (easily computable) lengths, the problem reduces to a standard length bash. $MF = BM - BF = \frac{c}{2} - (s - b) = \frac{c}{2} - \frac{a}{2} - \frac{b}{2} - \frac{c}{2} + b = \frac{b - a}{2}$ $MW = BM - BW = \frac{c}{2} - \frac{ac}{c + b} = \frac{ac + bc - 2ac}{2 (a + b)} = \frac{bc - ac}{2 (a+b)} = \frac{c}{a + b} (\frac{b - a}{2})$ $ML = BM - BL = \frac{c}{2} - \frac{a^2 - b^2 + c^2}{2c} = \frac{b^2 - a^2}{2c} = \frac{(b - a) (a + b)}{2c}$ Hence we compute $MW \cdot ML = \frac{(b - a)^2}{4} = MF^2$ as desired.