perform an inversion with arbitrary radius and center $C$(as usual the image $X$ is $X^{'}$) $\omega$ maps to a circle with diameter $P^{'}Q^{'}$ circle $k$ maps to a line tangent to $\omega^{'}$. $p(A,B)$ maps to a circle that is tangent to $k^{'}$ and whose center is on $P^{'}Q^{'}$ from here we deduce that $\omega^{'}$ ,$p(A,B)^{'}$ have equal radius on from here $\angle A^{'}P^{'}C^{'}=\angle A^{'}B^{'}C$ hence the result.

nikolapavlovic wrote: perform an inversion with arbitrary radius and center $A$(as usual the image $X$ is $X^{'}$) $\omega$ maps to a circle with diameter $P^{'}Q^{'}$ circle $k$ maps to a line tangent to $\omega^{'}$. $p(A,B)$ maps to a circle that is tangent to $k^{'}$ and whose center is on $P^{'}Q^{'}$ from here we deduce that $\omega^{'}$ ,$p(A,B)^{'}$ have equal radius on from here $\angle A^{'}P^{'}C^{'}=\angle A^{'}B^{'}C$ hence the result. What is $k'$ ? And if you can share a diagram it can be very good.

well it s a image of circle $k$ specified in the problem I will try to post a picture

Dear Mathlinkers, a synthetic proof without using inversion consists to use the Sawayama's theorem. Sincerely Jean-Louis

sorry guys can't you solve this problem without inversion or Sawayama theorem.Does it have normal solution or can someone post solution with inversion and picture please

IstekOlympiadTeam wrote: Let $\omega$ be the semicircle with diameter $PQ$. A circle $k$ is tangent internally to $\omega$ and to segment $PQ$ at $C$. Let $AB$ be the tangent to $K$ perpendicular to $PQ$, with $A$ on $\omega$ and $B$ on the segment $CQ$. Show that $AC$ bisects angle $\angle PAB$ Nice but very known problem with many solutions.I will give a hint : If k touches ω at $ E$, then $E,D,Q$ are on the same line($AB$ touches k at $D$).This needs some effort(homothety for example). But $PEDB$ is cyclic, hence : $QA^2=QB\cdot QP=QD\cdot QE=QC^2$ This gives $QA=QC$ and the rest of the problem is easy. Babis