Answer:
Given, TP and TQ are the tangents drawn on a circle with centre O. To prove: OT is the right bisector of PQ. Proof: In \[\Delta \text{ }TPM\]and \[\Delta \,TQM\] \[TP=TQ\] (Tangents drawn from external point are equal) \[TM=TM\] (Common) \[\angle PTM=\angle QTM\] (TP and TQ are equally inclined to OT) \[\therefore \Delta TPM\cong \Delta TQM\] (By SAS congruence) \[\therefore PM=MQ\] and \[\angle PMT=\angle QMT\] (By CPCT) Since, PMQ is a straight line, then \[\angle PMT+\angle QMT=180{}^\circ \] \[\therefore \angle PMT=\angle QMT=90{}^\circ \] \[\therefore \] OT is the right bisector of PQ. Hence Proved.
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