Let $\Omega$ be the circumcircle of the triangle $ABC$. The circle $\omega$ is tangent to the sides $AC$ and $BC$, and it is internally tangent to the circle $\Omega$ at the point $P$. A line parallel to $AB$ intersecting the interior of triangle $ABC$ is tangent to $\omega$ at $Q$.
Prove that $\angle ACP = \angle QCB$.
In this solution https://artofproblemsolving.com/community/c618937h1648567_problem_831_egmo_20135 (last one)
I have a few things that i did not understand …
$1$) Why $\sqrt{ab}$ inversion around $C$ followed by a reflection over the $C$-angle bisector sends $\omega$ to the $C$-excircle ?
$2$) And why it sends $P$ to $X$ and why it solves problem ?
thankyou very much
Best Answer
Let $A'$ denote the inverse of $A$ after that inversion you mentioned.Then,line $BC$ will go to the circumcircle of $ABC$ of the circumcircle.(Since $BC$ does not pass through $A$) .$AB$ and $AC$ line remains there.SO,the inverse of $\omega$ will be tangent to line $AB,AC,BC$,which is simple the excircle.and tangency points would be inverse of one another implying $P$ goes to $X$