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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 ?

please someone help

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Your link is not working, however I think you're referring to https://artofproblemsolving.com/community/c6h259941p8298724

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.

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