Let $\Gamma$ be the circumcircle of $ABC$. Let the intersection points of the external bisector of $\angle BAC$ with $BC$ be $T$, with $\Gamma$ be $R \neq A$ and with $\omega$ be $P$. Then, $AL$ and $PM$ are parallel since both are perpendicular to $AT$, which implies $TK/TM = TA/TP$. Looking at the power of $T$ with respect to $\omega$ and $\Gamma$ we obtain $TB \cdot TC = TP^2 = TA \cdot TR$ therefore $TK/TM = TA/TP = TP/TR$, hence $PK \parallel RM$. $RM$ is the perpendicular bisector of the side $BC$ so $PK \perp BC$ and $PKLM$ is a parallelogram with $PK = ML$. Let $Q$ be the reflection of $P$ with respect to $BC$, we will show that two circles are tangent to each other at $Q$. $BC$ is a diameter of $\omega$ hence $Q$ lies on $\omega$. $QK = PK = ML$ and both $QK, ML$ are perpendicular to $BC$ hence $KQLM$ is a rectangle, so $Q$ lies on the circumcircle of $KLM$ and it lies on the line passing through the center of both circles, which implies two circles must be tangent to each other at $Q$.