Intouch points of mixtilinear incircle and incenter are collinear proof
The A-mixtilinear incircle is defined to be the circle tangent to segments and of triangle and internally tangent to the circumcircle of . and denote the tangency points of the mixtilinear incircle with and respectively.
This is a (relatively) simple proof that points ,, and are collinear without the need for root-bc inversions or other transformations. Those are apparently more instructive though so you probably miss a few cool things.
Denote to be the center of the mixtilinear incircle, and to be the center of the .
,, and are collinear because both circles are tangent to eachother (tangent line at is perpendicular to both and ).
Consider the dilation centered at that sends to . This will take any point on the mixtilinear incircle to a point on because , and ratios of lengths are preserved so so lies on .
Let be the point that goes to. The line is perpendicular to before the transformation because is a tangency point of the mixtilinear incircle. is parallel to , so is also perpendicular to . This means that is the midpoint of arc because will be the perpendicular bisector of . That also means is the intersection of with . And , defined similarly is the intersection of with .
Now at this point we could just invoke pascal's theorem and call it done. But three of our points are determined by the other three, so it seems unnecessary that we should have to use pascal's theorem.
Since is similar to , , so we get that if is the intersection of line with , then is cyclic. By a similar argument we get that if is the intersection of line with , then is cyclic.
Now observe that can be seen as a miquel point of triangle . This means that and intersect at the same point (, so is cyclic)
So
But we said that lies on line , and lies on line , and the only way for this to happen is if .
As a corollary we also get that and are cyclic.