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Let $k$ = $k$($O$, $r$) be circumscribed circle for triangle Δ$ABC$ and let $m$ be perpendicular bisector of segment $BC$. Notice points: $m$ ∩ $p(A, C)$ = {$Q$}, $m$ ∩ $p(A, B)$ = {$P$} and let $ψ$k be inversion with regard to circle $k$. Prove that $ψ$k($P$) = $Q$.

So far I've deduced that $O$, $P$ and $Q$ are collinear points and noticed the point $T$, which is the intersection point of a tangent from $P$ on $k$. I've spent several days on this problem, trying to prove that $TQ$ ⊥ $p(P,Q)$ by using power of a point or by using four, basic, inversion properties. I've seen at least three other problems regarding this idea, but I was not able to solve any of them. I would, indeed, appreciate any help.

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    What does $p(A,C)$ mean? – 2017-01-08
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    I apologize for that. I was not aware it, might, not be standard notation. $p(A, C)$ is the line through points $A$ and $C$. – 2017-01-08
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    Thanks! Also, when you write "tangent from $P$ on $k$" you are implicitly assuming that $P$ is outside the circumcircle, which confused me initially (though certainly one of $P$ and $Q$ is). – 2017-01-08

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HINT.

Angles $\angle OAQ$ and $\angle OPA$ are equal. Thus triangles $OAQ$ and $OPA$ are similar and $OP\cdot OQ=OA^2$.

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    Thank you. It was, probably, simple for you, but very instructive for me. – 2017-01-08