Please
Q. In figure 10, $\angle$QPR = $\angle$PQR and M and N are respectively points on sides QR and PR of $△$PQR, such that QM = PN. Prove that OP = OQ, where O is the point of intersection of PM and QN.

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$$\begin{gathered} Given: \angle QPR = \angle PQR, QM = PN \\ To Prove: OP = OQ \\ Proof: In ∆PQR, \\ \angle QPR = \angle PQR \\ \Rightarrow RP = RQ [Sides opposite equal angles of a triangle are equal to each other] ...\left(i\right) \\ \Rightarrow \cancel{PN} + NR = RM + \cancel{MQ} [\because QM = PN] \\ \Rightarrow NR = RM ...\left(ii\right) \\ In ∆RQN and ∆RPM, \\ \angle QRN = \angle PRM \left[Common\right] \\ RP = RQ [From \left(i\right)] \\ RM = NR [From \left(ii\right)] \\ \therefore By SAS congruence criterion, \\ ∆RQN \cong ∆RPM \\ \because corresponding parts of congruent triangles are congruent \left(cpct\right), \\ \therefore \angle RPM = \angle RQN \\ \Rightarrow \cancel{\angle RPQ} - \angle MPQ = \cancel{\angle RQP} - \angle NQP [\because \angle RPQ = \angle RQP] \\ \Rightarrow \angle MPQ = \angle NQP ...\left(iii\right) \\ \therefore In ∆ OPQ, \\ \angle OPQ = \angle OQP [From \left(iii\right)] \\ \Rightarrow OP = OQ [\because Sides opposite equal angles of a triangle are equal] \\ Hence, proved \end{gathered}$$ 
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