A Self Posed Geo Prob

[Nirjhor]

Consider a circle $\omega$ with center $O$. A chord $AB$ subtends an acute angle at the center. Points $P$ and $Q$ are taken on segments $OA$ and $OB$ respectively, and two circles $\omega_1$ and $\omega_2$ are drawn with diameters $OP$ and $OQ$ respectively. Suppose $\omega_1\cap OB=M$ and $\omega_2\cap OA=N$. Prove that the diameter of $\bigcirc OMN$ is the radical axis of $\omega_1$ and $\omega_2$.

[Fm Jakaria]

Triangle $OPQ$ has foot of perpendiculars from $P,Q$ to be $M,N$. Letting the other foot from $O$ be $K$, and orthocenter $H$; the desired radical axis is $OHK$ and the diameter of $OMN$ is simply $OH$. Isn't that?

[Nirjhor]

Yes, I noticed later that the problem is trivial. Actually I started with a triangle with three altitudes, and inverted everything wrt a vertex as center. So my intended solution was: invert everything wrt $\omega$, then the problem becomes to prove that altitudes of a triangle concur.

Missed that a direct+easy solution is possible.