Question

If Rp and RQ are two tangents drawn from a point R to a circle with centre O, touching it at p and Q respectively, prove or is the perpendicular bisector of PQ

Answer 1

Answer:

Step-by-step explanation:

Consider RP and RQ are two tangents drawn from a point R to a circle with centre O, touching it at P and Q respectively,

Draw a line between PQ which intersect OR at the point M

Prove that $PM=QM$ and $OR\perp PQ$

In $\Delta ORP\text{ and }\Delta ORQ$

$OP = OQ\text{ (radius of the circle)}$

$RP=RQ\text{ (equal tangent segments)}$

[if two tangents are drawn to a circle, from one external point, then they have equal tangent segments]

$OR\text{ (common side)}$

Therefore, by SSS criterion of congruence, $\Delta ORP\text{ and }\Delta ORQ$  are congruent.

So, $\angle POR=\angle QOR$

therefore, $\angle POM=\angle QOM$  because, M is the point of intersection of OR and PQ

In $\Delta OMP\text{ and }\Delta OMQ$  

$OP = OQ\text{ (radius of the circle)}$

$\angle POM=\angle QOM\text{ (proved earlier)}$

$OM\text{ (common side)}$

Therefore, by SAS criterion of congruency,  $\Delta OMP\text{ and }\Delta OMQ$  are congruent.

So, PM = QM

$\Delta PMO\text{ and }\Delta QMO$  

Now,

$\angle PMO+\angle QMO=180^{\circ}\\\\\therefore2(\angle PMO)=180^{\circ}\text{  }[\angle PMO=\angle QMO]\\\\\therefore\angle PMO=90^{\circ}$

Thus, PM=QM and $\angle PMO=\angle QMO=90^{\circ}$

Hence, OR is the perpendicular bisector of PQ.