Prove that LM is parallel to AB the question is the following:

The problem is equivalent to show that $M$ is a midpoint of arc $\widehat{AB}$. I made a diagram on geogebra (points $C$ and $P$ got confused, sorry):

It is not hard to prove $AMYX$ is cyclic. I was hoping to find some congruent triangles that would help, but it didn't quite work: $AM=MB$ and $MY=MT$ so I put my hope of finding congruence there. I didn't want to work with the second meeting of $PB$ and the left circle, because it seems unnecessary. Got a lot of stuff going on here: $C$ is the circumcenter of $\triangle MTY$, $AT \perp TX$
EDIT: we got a Protassov configuration! I think the incenter of $\triangle ABC$ (in the top image, $\triangle ABP$ in the bottom) has a nice alignment with $T$ and a midpoint of arc $AB$
 A: Mirror the blue circle in the line $LM$. Then the line $BC$ is simply another common tangent line of two circles. In other words, we are really trying to show that the dashed line defined by the red lines in the picture below is parallel to the symmetry axis.
The key point is to show that two of the common tangent lines (the green and one of the reds in the picture below) intersect on the circle.

The first step is to apply circle inversion with respect to the dashed circle in the picture below. I have indicated the corresponding old and new objects with like colours.

We are left to show that the green circle, defined to touch one of the lines and one of the blue circles while going through the point of symmetry, touches the line in precisely the intersection point of the red circle with the line.
I have included a clearer picture below and gave some points names.
Starting with the rectangle, construct the red circle and hence the point $X$. Let $P$ be the projection of $B$ on $XC$ and define the green circle to be the circumcircle of $X$, $P$ and $M$. By Thales, to show that this satisfies all we want of the green circle, it suffices to show equality between the blue angles.

The proof is surprisingly short. Since $YD$ touches the red circle, we have $|YD|^2 = |YM||YX|$. By symmetry $|YD| = |XB|$, so we even have that $|XB|^2 = |XM||XY|$.
From the similarity between triangles $XPB$ and $XBC$, we get that $|XB|^2 = |XP||XC|$. Combining this with the previous, gives us that $|XP||XC| = |XM||XY|$. In other words, we have shown that $YMPC$ are all on a circle. We immediately find $\angle BXC = \angle YCP = \angle PMX$, as was to be shown.
A: Given two circles with centers $K$, $L$, externally tangent at $E$, with common tangent $AH$ and tangent $LM$, join $MA$, make chord $MB=MA$, and draw $BD$ tangent to the other circle.
Since triangle $ABM$ is isosceles, then $\angle BAM=\angle MBA$. And since $\angle LMA$ between chord and tangent is equal to $\angle MBA$ in the alternate segment, then$$\angle LMA=\angle BAM$$Therefore$$LM\parallel AB$$
Note: On this interpretation of the question and its conditions, tangents $AH$ and $BD$ are extraneous.  But even if the tangents are included, and $A$ is therefore not random, $B$ cannot be supposed random, but must be chosen such that triangle $ABM$ is isosceles. In other words, $M$ is the midpoint of arc $AMB$. Otherwise $LM$, which is given in position for the given circles, will not be parallel to $AB$. A fuller statement of the conditions in this question would be helpful.

Addendum:
In light of the clarifying comments, that the above solution misinterprets the question, let me offer at least a proof for the special case of equal circles.
In the figure below, with $F$ as the intersection of the tangent at $E$ with tangent $LM$, it is clear by symmetry that tangent $CBFD$ passes through center $K$.
Since hypotenuse $KL=2KM$, then $\angle MKL=60^o$.
And since $\angle FKL=\angle KLF=30^o$, therefore exterior $\angle KFM=60^o$.
Further, since $\angle AKB= 90^o-30^o=60^o$, then $\triangle AKB$ is equilateral, $$\angle ABK=60^o=\angle KFM$$and$$LM\parallel AB$$And it is clear in this case that $M$ is the midpoint of arc $AMB$.

