# In figure 3, two Given : Two tangents RQ and RP from an external point R to the circle with center O and

PRQ = 120°

To Prove : OR = PR + QR

Construction : Join OP and OQ

Proof :

PR = RQ [ Tangents drawn from an external point to a circle are equal] …

Also OP PR and OQ RQ

[ as tangent drawn at a point on a circle is perpendicular to the radius through the point of contact]

OPR = 90°

OQR = 90°

OPR + PRQ + OQR + QOP = 360°

90 + 120 + 90 + QOP = 360

QOP = 60°

In OPR and ORQ

OP = OQ [radii of same circle]

PR = RQ [By 1]

OR = OR [common]

OPR ORQ [By SAS]

POR = ROQ [By CPCT]

QOP = POR + ROQ

POR + POR = 60°

POR = 30°

As ROP is a right angled triangle

sin POR = PR/OR

sin 30° = PR/OR

1/2 = PR/OR

OR = 2PR

= PR + PR = PR + QR [from 1]

Hence Proved.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 