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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?

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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:

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

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  • $\begingroup$ Wouldn't $$\angle (t,SR)=\angle SRM$$ $\endgroup$ – wesdrxvrtgf Sep 24 '18 at 16:32
  • $\begingroup$ Why? I don't think so. $∠(t,SR)=∠SMR$ is because tangent chord property $\endgroup$ – Aqua Sep 24 '18 at 16:34
  • $\begingroup$ Well since the alternate segment theorem, states that the angle between the tangent and chord at the point of contact is equal to the angle in the alternate segment then the angle between the tangent and chord is $$\angle (t,SR) and $$\angle SRM is the angle in the alternate segment they be equal as far as I am aware $\endgroup$ – wesdrxvrtgf Sep 24 '18 at 16:38
  • $\begingroup$ Yes, and that is $\angle SMR$ $\endgroup$ – Aqua Sep 24 '18 at 16:40
  • $\begingroup$ Oh right got it $\endgroup$ – wesdrxvrtgf Sep 24 '18 at 16:41
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You are right about “alternate segment theory” is needed in solving this problem. However, the $O_1$ and $O_2$ are distractions.

enter image description here

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”).

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