Let $D$ be the given tangency point. Let $N$ be the reflection of $M$ with respect to the midpoint of $[KL]$. Let $L'$ be the intersection of the external bisector of angle $BAC$ with $(ABC)$. Note that $L, M$ and $L'$ are collinear, $KMLN$ is a rectangle and $MD \parallel AL$. Then, by similarity and PoP, we get $$\frac{MD}{L'M}=\frac{AL}{L'L}=\frac{ML}{MD}$$Since $\frac{AL}{L'L}=\cos \angle KLM = \frac{ML}{KL}$, we get $KL=MN=MD$. It can be easily seen that $N$ is the desired tangency point.

If $AB$ were equal to $AC$, $K$ would coincide with $M$, and $KLM$ wouldn't be a triangle. Therefore, the external angle bisector of $\angle BAC$ intersects $BC$ at a point, say $X$. Denoting the circle with diameter $[BC]$ by $(BC)$, let $AX$ touch $(BC)$ at $T$, and let $PX$ be the other tangent line from $X$ to $(BC)$ with $P \in (BC)$. By the Apollonian circle theorem, we have $(B, C; X, K) = -1$. On the other hand, since $BTCP$ is a harmonic quadrilateral, we know that $(B, C; T, P) = -1$. Taking perspectivity from $T$ to $BC$, we obtain $(B, C; X, TP \cap BC) = -1$, which implies $TP \cap BC = K$. Since $TK$ and $LM$ are perpendicular to $BC$, and $MT$ and $KL$ are perpendicular to $AX$, $TKLM$ is a parallelogram. In particular, $ML = TK = KP$. Since $KP$ and $ML$ are perpendicular to $BC$, thus parallel, and $KP = ML$, we know that $KPLM$ is a rectangle. It follows that $P \in (KLM)$. Let $L'$ be the second intersection point of $AX$ and $(ABC)$, which is the midpoint of the arc $BAC$. Since $\angle KAL' = \angle KML'$, we know that $AKML'$ is a cyclic quadrilateral. The power of a point theorem implies $XK \cdot XM = XA \cdot XL' = XB \cdot XC = XP^2$. Therefore, $XP$ is tangent to both $(KLM)$ and $(BC)$ at point $P$.

Official solution: Denoting the circle with diameter $[BC]$ by $(BC)$, let the external angle bisector of $\angle BAC$ touch $(BC)$ at $P$, intersect $(ABC)$ at $R \neq A$, and intersect $BC$ at $T$. $AL$ and $PM$ are perpendicular to $AT$ and thus parallel. Besides, the power of a point theorem implies $TB \cdot TC = TP^2 = TA \cdot TR$. Therefore, $$\frac{TK}{TM} = \frac{TA}{TP} = \frac{TP}{TR}.$$It follows that $PK$ and $RM$ are parallel, which means $PK \perp BC$. Thus, $PKLM$ is a parallelogram and $PK = LM$. Let the reflection of $P$ across $BC$ be $Q$. By definition, $Q \in (BC)$. Furthermore, since $KQLM$ is a rectangle, we have $Q \in (KLM)$. The fact that $Q$ is on the line connecting the centers of the circles $(BC)$ and $(KLM)$ implies that these circles are tangent to each other at $Q$.

Wikiliks wrote: Let $L'$ be the second intersection point of $AX$ and $(ABC)$, which is the midpoint of the arc $BAC$. Since $\angle KAL' = \angle KML'$, we know that $AKML'$ is a cyclic quadrilateral. The power of a point theorem implies $XK \cdot XM = XA \cdot XL' = XB \cdot XC = XP^2$. Therefore, $XP$ is tangent to both $(KLM)$ and $(BC)$ at point $P$. This step is unnecessary, given that $P$ lies on $(BC)$, $(KLM)$, and the line connecting the center of these two circles.

Consider the inversion around $\omega$ and define $T$ as the tangency point between $\omega$ and the external bisector of $\angle{BAC}$. First of all, if we let $K'=AT\cap BC$, then $K'$ is the inverse of $K$ with respect to $\omega$: indeed, since $K$ and $K'$ are the feet of the internal and the external bisector of $\angle{BAC}$, we have $(K',K;B,C)=-1$, which implies that $MK\cdot MK'=BM^2$, as desired. This means that $K$ belongs to the circumference which is the inverse of line $AT$, that is, the circumference with diameter $MT$. Therefore, $\angle{MKT}=90^{\circ}$. We also know that $\angle{KML}=90^{\circ}$, so $KT\parallel ML$. Moreover, since $\angle{KAT}=\angle{MTA}=90^{\circ}$, we also have $AL \parallel TM$. Hence, $TKLM$ is a parallelogram, which gives $TK=ML$. Define now $D$ as the symmetrical point of $T$ with respect to $K$, so that $KD=TK=ML$. Consequently, $MLDK$ is a rectangle and $D\in (KML)$; furthermore, $MD$ is a diameter of this circumference. Also, because $K$ lies on the diameter of $\omega$ and $\angle{MKT}=90^{\circ}$, we have that $D\in \omega$ as well. This implies that $D$ is the tangency point of $\omega$ and $(KML)$, as desired.

Attachments:

Denote $N$ as the midpoint of the arc $BAC$ and let $T$ be the point where $(BC)$ is tangent to $NA$. Take the inversion centered at $M$ with radius $MB\sqrt{-1}$. $B\leftrightarrow C, N\leftrightarrow L, T \leftrightarrow T^*,A\leftrightarrow A^*$ $N,A,K,M$ are cyclic so $L,A^*,K^*$ are collinear. $\angle KNL=\angle K^*KL=180-\angle AKM=\angle MNA$ Let $U$ be the altitude from $M$ to $NK^*$. $\angle MNT=\angle UNM$ so $MNT\cong MNU$ and this gives that $MU=MT$ and $MU\perp K^*N$ hence $NU$ is tangent to $(BC)$. $(BC)\rightarrow (BC), (KLM)\rightarrow K^*N$ $K^*N$ is tangent to $(BC)\implies (KLM)$ and $(BC)$ are tangent as desired.$\blacksquare$

Solved with Shreyasharma. Why did we struggle unnecessarily on this? Let $T$ be the midpoint of arc $BAC$ (containing $A$) and $X$ be the point of tangency of $AT$ and $(BC)$. We first make the following useful observation. Claim : $XM \parallel AL$ Proof : Simply note that since $M$ is the center of $(BC)$ we have $XM \perp AT$. Further, it is well known (and trivial to show) that $AL \perp AT$ ($T$ and $L$ are antipodal points). Thus, it follows that $AL \parallel XM$. Now, comes the key claim. Claim : The circles $(KML)$ and $(BC)$ intersect at a point $P$ which lies on $\overline{XK}$. Proof : Let $P'=\overline{XK} \cap (KML)$. Note that, \[\measuredangle XP'L = \measuredangle KP'L = \measuredangle KML = 90^\circ = \measuredangle TAL = \measuredangle XAL\]Thus, $XAP'L$ is cyclic. From this it follows that \[BK \cdot KC = KA \cdot KL = XK \cdot KP'\]which concludes that $P'$ also lies on $(BC)$ as desired. Now, we prove the following claim which finishes the problem. Claim : $XP \perp BC$. Proof : Simply note that since $P$ lies on both $(BC)$ and $(KLM)$ by the previous claim, we have \[\measuredangle PXM = \measuredangle XPM = \measuredangle KPM = \measuredangle KLM = \measuredangle ALT = \measuredangle XMT \]Thus, it is clear that $XP \parallel TL$. Since $TL \perp BC$ (well known that $TL$ is the perpendicular bisector of $\overline{BC}$) it follows that $XP \perp BC$ as desired. Now, the finish is clear. Simply note that by the previous perpendicularity, it follows that $KPLM$ is a rectangle. So, its center ($O$) is the midpoint of $PM$. Thus, the centers of the circles $(KLM)$ and $BC$, are collinear with $P$ ($P-O-M$) from which the desired tangency follows.

Let the tangent point on ecternal bisector of $\angle{BAC}$ be $T$. Its easy to see $\angle{LAT}=\angle{MTA}=90^\circ$ and $MT||AL$. Let $MT\cap AC=R$. By $MT||AL$ and angle chasing we can say $\triangle {BKL}\sim\triangle {RCM}$ and if the similarity ratio is equals to $1$, this means $|KL|=|MC|$ and the radius of $(KLM)$ equals to half of $BC$-diameter circles radius. And symmetry of $M$ to $KL$'s middle is on $BC$-diameter circle. And we will done. Lets prove the similarity ratio is equal to $1$. Let $|BL|,|BK|,|KL|$ be $x,y,z$ respectively. Let the similarity ratio be $k$. We can say $|RC|=xk$, $|RM|=yk$, $|CM|=zk$,$|TR|=zk-yk$,$|KM|=zk-y$. $$\triangle {RCM}\sim \triangle {ACK}\Rightarrow|AR|= \frac{x(zk-y)}{z}$$$$\triangle {BLM}\sim\triangle {RAT}\Rightarrow |AR|=\frac{x(z-y)}{z}$$. We are done. $\square$