https://www.cut-the-knot.org/Curriculum/Geometry/FourCrossingCircles.shtml

I started calling c the center of the concentric circles I realized that AcB is equal to A2cB2 and A1cB1 since the three triangles are equals since theirs sides equals, so by rotating with center in c such angle A1cB1, points A1 and A2 will be on the radii cB1 and cB2 respectively, therefore chords A1A2 and B1B2 have to be proportional to Ra and Rb

Ah, so I guessed.

Could you please have a look at

https://www.cut-the-knot.org/Curriculum/Geometry/CirclesOnPerpBisector.shtml

and

https://www.cut-the-knot.org/Curriculum/Geometry/ParallelToBase.shtml

If there is a simple solution it escapes me.

Thank you.

Dear Alex,
I think we can generalize the fact as following:

Given two circles (O1), (O2) with two external tangent lines eT1, eT2 and two internal tangent lines iT1, iT2. The circle (O) is tangent with both (O1) and (O2).

1. Both internal tangent case:
If (O) is internal tangent with (O1) and (O2) then (O) cuts two internal tangent lines iT1, iT2 at four points and there are two lines (each connected two from these four points) parallel with two external tangent lines eT1, eT2.

2. One internal, one external tangent case:
If (O) is external tangent with (O1) and internal tangent with (O2) then (O) cuts two external tangent lines eT1, eT2 at four points and there are two lines (each connected two from these four points) parallel with two internal tangent lines iT1, iT2.

I think you can express the facts in better ways.
I stil not found proofs for these facts.
Best regards,
Bui Quang Tuan

Thank you for looking into that.

> I think we can generalize the fact as following

You are very likely right. I still do not have any idea of how to approach these problems.

All the best,
Alex

I submit to your always helpful review my approach to Mr. J. Marshall Unger Theorem, since we, the engineers, tend to be too much simplistic,... and sometime simplicity it can be deceiving.

The approach It is based on Feuerbach Theorem and Inversion.

https://www.cut-the-knot.org/Curriculum/Geometry/FeuerbachProof.shtml

To clarify, I summarize the rationality of the attached sketch, somewhat untidy, my old CAD software it just like me, too old and ill fitted for Geometry.

1) We are given to circles ( I ) and (P) the two common exterior tangents concurrent at A

2) We draw just one of the other two common interior tangents to keep clear the drawing

3) We draw also a parallel to one of the interior tangent that will be tangent to circle (P).
Using same notation we will call B and C the intersection of this parallel with the original exterior tangents.

4) So we end having the a triangle ABC in which ( I ) is the incircle, and we will have another Triangle AMN in which ( P ) will be its incircle

5) If we draw the Feuerbach circle for the AMN triangle. The resulting cicle will be tangent to circles ( I ) and (P) as per Feuerbach Theorem

6) Since Feuerbach circle cut the exterior tangent at midpoints of the sides AM and AN we called F1 and F2 line F1F2 is parallel to interior tangent MN.

Unfortunately our drawn Feuerbach circle it is not tangent to the circles (I) (internally) and (P) (externally) but on the reverse.

7) To solve that we will made an inversion, taking A as the center of inversion and radius of inversion such, that the inversion circle cut orthogonally our Feuerbach circle, so it will remain unaltered.

8) Now after inversion circles ( I ) and (P) will became circles ( i ) an ( p ), which are well known homothetic to ( I ) and (P) with same homothecy ratio and center A. and will be both tangent to the unaltered Feuerbach circle.

Therefore by the properties of the homothecy the new interior tangent mn which obviously remain parallel to F1F2 and to BC

could you please supply a link to Marshall Ungar's theorem. Just tp save time.

Thank you,
Alex

I took from your reference link at the bottom post #8 of this subject.

References

1. J. Marshall Unger, A new proof of a "hard but important" Sangaku problem, Forum Geometricorum, 10 (2010) 7--13.

Anyway my proof I think need your always helping hand to fulfill the mathematics standards.

could you please provide a link to "this subject".

https://forumgeom.fau.edu/FG2010volume10/FG201002.pdf
page 8 Theorem 1

and from
https://www.cut-the-knot.org/Curriculum/Geometry/CirclesOnPerpBisector.shtml
https://www.cut-the-knot.org/Curriculum/Geometry/ParallelToBase.shtml#Unger

I thought a case where to use the Feuerbach Theorem

I am sure Feuerbach theorem is quite relevant to the problem. However, the resulting circle must path through B and C. This, probably, follows because you deal with homotheties. Very likely. With this caveat, there is another point: what you showed is that if one of the common internal tangents is parallel to the base there is circle (O). Now we also need a converse.