Let $ ABC$ be a triangle with circumcentre $ O$. The points $ P$ and $ Q$ are interior points of the sides $ CA$ and $ AB$ respectively. Let $ K,L$ and $ M$ be the midpoints of the segments $ BP,CQ$ and $ PQ$. respectively, and let $ \Gamma$ be the circle passing through $ K,L$ and $ M$. Suppose that the line $ PQ$ is tangent to the circle $ \Gamma$. Prove that $ OP = OQ.$
My idea to solve this was a following:
First we prove converse, that is if $OP = OQ$ then $\Gamma$ is tangent to $PQ$.
Now to the problem. Fix $Q$ on $AB$ and let us move $P$ on a line $AC$. Then $L$ is fixed and $M$ move on a parallel to $AC$ through $L$ so this parallel, (name it $\ell$) is also fixed. Point $K$ also move but on midlle line parallel to $AC$. Clearly $M$ and $K$ depend only on $P$. Now observe the function $$f(P) = \angle (KL,\ell) -\angle (QP,AB)\;\;\; (=\alpha-\beta)$$
I would like to prove that as $P$ moves $f(P)$ is only twice $0$ and since converse is true (in those two cases) then we are done.
My problem is: how to formalise $f(P)$?