We have an $\angle POQ=120^\circ$ and a point $M$ that is inside the angle. Perpendiculars are drawn from $M$ to the arms and the angle bisector of $\angle POQ$. Show that $\triangle ABC$ is equilateral.
From $AOBM \Rightarrow \angle AMB=60^\circ$ and from $OBMC \Rightarrow \angle BMC=120^\circ$, thus $\angle AMB=\angle AMC=60^\circ$. We notice $OBMC$ is inscribed. Let the center of the circumscribed circle $K(K; R)$ is $K$. Now, I am trying to show that $A$ lies on $K\Leftrightarrow AK=KB=R$. I don't see how to make it! Take you in advance!
Hint: Show that $A, B, C$ lie on a circle with diameter $OM$.
Hint: $\angle ABC = \angle AOC$.
Can you complete it from here?
With regards to your observation that "We notice $OBMC$ is inscribed", another way to continue from there is to show similarly that $OACM$ is a cyclic quad. Since the circumcircle of $OCM$ is uniquely determined, and must be the cyclic quad in both cases, hence $OACMB$ is concyclic.