Proposition1(Sawayama): Circle is internally tangent to circle at . are two chords of that are tangent to at respectively. Prove that the incenters of lie on .

Lemma1: If is constant for , and is tangent to at .

Let . where is the radius of circle , and this proves our lemma.

Correlary1: bisects by the converse of angle bisector theorem: .

Proof of our proposition: Let denote the incenter of , . By Correlary1 are collinear since is the midpoint of arc . By similarity, , which is the constant described in Lemma1, therefore , the power of wrt . Since , thus and . Since , therefore . The last angle equality implies are collinear and we are done because the incenter of lies on by symmetry.

Some properties I want to highlight and are trivial from the proof above:
Let denote the incenter of .
1. and are orthogonal
2. are cyclic quadrilaterals.
3.
4. are isogonal wrt .
5. are concyclic.


Property6: is the midpoint of arc that doesn't contain . Prove that are concyclic.

Proof: By property3 are isogonal wrt . Since externally bisects , it also externally bisects . As by properties of incenters, therefore is cyclic.

Correlary2: are concurrent by radical axis theorem.

Property7: is the midpoint of arc that does not contain . Prove that are collinear. This is immediate by pascal's theorem.

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Proposition 2: are internally tangent to at . are chords of that are also internal common tangents of like so in the diagram. Prove that the external tangent of these two circles closer to than is parallel to .

Proof:

Let be the midpoint of arc that doesn't contain . Define and similarly. Thus is antiparallel to wrt . It suffices to show that , which implies is the external tangent. We will show that .

is tangent to at , is tangent to the same circles at . By property7 we know are collinear. Thus . In addition, . Dividing these two gives us . Since , the analogous expression for equals and we are done.