Let *I* be the incenter of the triangle *ABC*. Let $\Gamma$ denote the circle passing through $A$, $K$, $L$. We have
$$\begin{align*} \angle AKD = \angle EKD &= 180^\circ - \angle DME = \angle DMF = 180^\circ - \angle FLD = \\ &= 180^\circ - \angle ALD. \end{align*}$$
Therefore, $D$ lies on $\Gamma$. So, to prove that $\Gamma$ touches $BC$ it suffices to show that $\angle DAL = \angle BDL$.

Since $AE = AF$ and $EM = MF$, it follows that $AM \perp EF$ and $AM$ is the bisector of the angle $EAF$, i.e., the points $A$, $M$, $I$ lie on the same line.
We have
$$\begin{align*} \angle BDL &= 90^\circ - \angle IDL = 90^\circ - (\angle IDM + \angle MDL) = \\ &= (90^\circ - \angle MDL) - \angle IDM = [\angle MDL = 180^\circ - \angle MFL = \angle MFA] = \\ &= (90^\circ - \angle MFA) - \angle IDM = \angle FAM - \angle IDM = \\ &= \angle DAL + \angle DAM - \angle IDM. \tag{1} \end{align*}$$

Since the triangle $AFI$ is the right-angled triangle and $FM \perp AI$, we have $IF^2 = IM \cdot IA$. Since $IF = ID$, we obtain $ID^2 = IM \cdot IA$. By the Power of a Point Theorem, the circle passing through the points $A$, $M$, $D$ touches the line $ID$ at $D$, then $\angle IDM = \angle DAM$. From (1) it follows that $\angle DAL = \angle BDL$ as required.