Geometry problem solving involving alternate segment theory

Question

In the diagram below, prove that: $$\angle QMP=\angle RMP$$ .

I am pretty sure that we need to use the alternate segment theory here but I am not sure how?

Answer 1

Let $H$ be a homothety which takes smaller circle to bigger. Then it takes $P$ to a new point $S$ on bigger circle and line $PQ$ to parellel line $t$ through $S$, so $t$ is tangent on bigger circle. Now we have:

$$\angle QMS = \angle SRQ = \angle (t,SR) = \angle SMR$$

Answer 2

You are right about “alternate segment theory” is needed in solving this problem. However, the $O_1$ and $O_2$ are distractions.

Draw the tangent at M and let it cut QPR extended at T. By tangent properties, the blue marked angles are equal (= x, say).

$\angle PMR = x – z$

$= (\angle Q + y) – z$, .... (exterior angle of triangle)

$= y$ .... (because $\angle Q = z$, by “angles in alternate segment”).