Two circles touch internally at M. A straight line touches the inner circle at P and cuts the outer circle at Q and R. Prove that Angle QMP = Angle RMP..

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$$\begin{gathered} \mathrm{Construction}:\mathrm{Draw} \mathrm{a} \mathrm{tangent} \mathrm{to} \mathrm{the} \mathrm{circle} \mathrm{at} \mathrm{their} \mathrm{point} \mathrm{of} \mathrm{contact} \\ \mathrm{Solution}: \\ \mathrm{Let} \angle \mathrm{SMQ} = \mathrm{y} \\ \mathrm{then}\angle \mathrm{SMQ} = \angle \mathrm{MRQ} \left( \mathrm{Alternate} \mathrm{segment} \mathrm{theorem}\right) \\ \mathrm{Let} \angle \mathrm{TPR} = \mathrm{x} \\ \mathrm{then} \angle \mathrm{TPR} = \angle \mathrm{TMP}= \mathrm{x} \left( \mathrm{Alternate} \mathrm{segment} \mathrm{theorem}\right).....\left(1\right) \\ \angle \mathrm{MPQ} = \mathrm{x} + \mathrm{y} \left( \mathrm{Exterior} \mathrm{angle} \mathrm{of} \mathrm{\Delta } \mathrm{RMP}\right) \\ \angle \mathrm{MPQ} = \angle \mathrm{MTP} = \mathrm{x} + \mathrm{y} \left(\mathrm{Alternate} \mathrm{segment} \mathrm{theorem}\right) \\ \angle \mathrm{MTP} = \angle \mathrm{PMS} = \mathrm{x}+ \mathrm{y} \left( \mathrm{Alternate} \mathrm{segment} \mathrm{theorem}\right) \\ \therefore \angle \mathrm{QMP} = \angle \mathrm{SMP} -\angle \mathrm{SMQ}= \mathrm{x}+ \mathrm{y} –\mathrm{y}= \mathrm{x}.......\left(2\right) \\ \mathrm{From} 1 \mathrm{and} 2 \mathrm{we} \mathrm{get} \\ \angle \mathrm{TMP} = \mathrm{QMP} = \mathrm{x} \\ \mathrm{Hence} \mathrm{proved}..... \end{gathered}$$

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