I don't know how to draw here, so draw the figure on paper.

Draw two circles (no need to be same in size) that kiss each other at point A.

Call the center of one circle, B, and the center of the other circle, C. Draw line segment BC.

Draw line segment PQ such that it passes through point A, it ends on the other side of circle B at point P, and it ends also at the other side of circle C at point Q.

Draw radii BP and CQ.

Draw the tangent lines at points P and Q.

Extend radius PB to meet the (tangent line at Q) at point R.

Extend radius QC to meet the (tangent line at P) at point S.

Now we are set.

Angles BAP and CAQ are equal, because they are vertical angles. ----(1)

Triangle PBA is isosceles, because radius BP = radius BA.

So, angle BPA = angle BAP ------------------------------------(2)

Triangle QCA is isosceles, because radius CQ = radius CA.

So, angle CQA= angle CAQ ------------------------------------(3)

Play with (1),(2),(3),

Hence, angle BPA = angle CQA ------------------(4)

In circle B:

radius BP is perpendicular to the tangent line PS, so angle BPS is 90 degrees.

And so, angle APS = 90deg minus angle BPA --------------(5)

In circle C:

radius CQ is perpendicular to the tangent line QR, so angle CQR is 90 degrees.

And so, angle AQR = 90deg minus angle CQA --------------(6)

Play with (4),(5),(6),

Hence, angle APS = angle AQR

Angles APS and AQR are alternate interior angles of tangents PS and QR, with line segment PQ as the transversal line.

Therefore, since angles APS and AQR are equal, then tangents PS and QR are parallel to each other. ----------proven.

------------------------------------------------------------

If you don't like that reasoning, we can show that angles PRQ and QSP are 90 degrees each. So PRQS is a rectangle. So RQ is parallel to PS.