question_answer

From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQ.

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.