$\angle KDL=\angle KDM+\angle LDM=\angle AFE+\angle AEF=180^\circ-\angle KAL$ so quadrilateral $KALD$ is inscribed in a circle. We have to prove $|BL|\cdot |BA |=|BD|^2$ which is equivalent to (tangent segments have equal length) $$|BL|\cdot |BA |=|BF|^2$$Let the inscribed circle be the unit circle on complex plane. With small letters we denote complex cordinates of points denoted with big letters. $$a=\frac{2ef}{e+f}\wedge b=\frac{2df}{d+f}\wedge 2m=e+f$$$L$ lies on line $BF$ so using collinearity criterion $$l=\frac{2f-l}{f^2}$$Quadrilateral $DMFL$ is cyclic so using concyclicity criterion and above equality $$\frac{f-e}{2d-e-f}\cdot\frac{d-l}{f-l}=\frac{f-m}{d-m}:\frac{f-l}{d-l}=\overline{\left(\frac{f-m}{d-m}:\frac{f-l}{d-l}\right)}=\frac{e-f}{2ef-de-df}\cdot\frac{2df-dl-f^2}{f-l}$$We can divide both sides through $\frac{f-e}{f-l}$. After transformations $$l=\frac{f^2+e(d+f)-3df}{2(e-d)}$$Hence $$b-a=\frac{2f^2}{d+f}\cdot\frac{d-e}{e+f}$$$$b-f=f\cdot \frac{d-f}{d+f}$$$$b-l=\frac{(e+f)(d-f)^2}{2(d-e)(d+f)}$$Thus $$(b-a)(b-l)=(b-f)^2$$which is at least as strong result as we need.

Let $X$ be the intersection point of $EF$ and $BC$ and $Y$ the intersection of $AD$ and $EF$. It is known that $(E,F;Y,X)$ are in harmonic division so $XE\times XF=XY\times XM$. But, $XE\times XF=XD^2$. Hence, the circumcircle of $YMD$ is tangent to the line $BC$ at point $D$ and also to the incircle. And we find the following relation with angles : $\widehat{YDB}=\widehat{YMD}$. By Miquel's theorem in triangle $AEF$ and with points $L$, $M$, $K$ we have $D$ on the circle $AKL$. So we have to prove that $AKL$ is tangent to $BC$ at $D$, or equivalently that $\widehat{BDA}=\widehat{AKD}$ but the last identity is true since $\widehat{ADB}=\widehat{FMD}=\widehat{AKD}$.

Let $DA$ and $DL$ intersect the incircle at $T$ and $P,$ respectively. Clearly, $DT$ is the $D$-symmedian in $\triangle{DFE} \implies$ $$\angle{DPT}=\angle{DFT}=\angle{DME}=\angle{DLA} \implies LA \mid \mid PT$$By Miquel on $\{M, L, K\}$ wrt $\triangle{AFE}$ we have that $ADLK,$ which by the above parallelism and homothety is tangent to the incircle and $BC,$ consequently.

Peru TST 2015 P2 wrote: The inscribed circle of the triangle $ABC$ is tangent to the sides $BC, AC$ and $AB$ at points $D, E$ and $F$, respectively. Let $M$ be the midpoint of $EF$. The circle circumscribed around the triangle $DMF$ intersects line $AB$ at $L$, the circle circumscribed around the triangle $DME$ intersects the line $AC$ at $K$. Prove that the circle circumscribed around the triangle $AKL$ is tangent to the line $BC$. Solution: $\angle LAK=180^{\circ} -\angle AFE-AEF=180^{\circ}-\angle LDM-\angle KDM=180^{\circ}-\angle LDK $ $\implies$ $LAKD$ is cyclic. Also, $\overline{DA}$, $\overline{DM}$ are isogonal WRT $\angle FDE$, Hence, $$\angle DLK=\angle DAE=180^{\circ}-\angle ADE-\angle DEA=180^{\circ}-\angle FDM-\angle DEA=\angle EDC-\angle FDM$$$$=\angle EDC-(180^{\circ}-\angle EKD-\angle KED)=\angle EDC-\angle EDK=\angle KDC$$Hence, $\odot (ALK)$ is tangent to $BC$ at $D$