# $2009$ Japan olympiad Problem $4$

Triangle ABC has circumcircle $$Γ$$, A circle with center $$O$$ touches to line segment $$BC$$ at $$P$$ and touches the arc $$BC$$ of $$\Gamma$$ which doesn't have $$A$$ at $$Q$$. If $$\angle {BAO} = \angle {CAO}$$, then prove that $$\angle {PAO} = \angle {QAO}$$

In this Solution ($$14$$th post) they have used a $$\sqrt{bc}$$ inversion centered at $$A$$ but where we have used that $$AO$$ is bisector ? i mean we can take any $$O$$ and draw a circle around it tangent to $$BC$$ and circumcircle say $$P$$ and $$Q$$ then now with this inversion we can again say that $$P$$ goes to $$Q$$ and $$Q$$ goes to $$P$$ ,so they are isogonal always ?

Call the circle having center at $$O$$ and passing through $$P$$ and $$Q$$ as $$\omega$$. Since $$AO$$ is the angle bisector, $$\sqrt{bc}$$-inversion and flip still preserves the line $$AO$$ (you cannot preserve the line $$AO$$ without it being the angle bisector). The rest is clear I think : the image of $$\omega$$ is still tangent to $$BC$$ and $$\Gamma$$, and since it has its center on the image of line $$AO$$ which is $$AO$$ itself, the center must be $$O$$, meaning $$\omega$$ is fixed. This means $$P$$ and $$Q$$ are mapped to each other.