In figure PA and PB are tangents from P to the circle with centre O. R is a point on the circle. Prove that : PC + CR = PD + DR
Since the tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠OPT = 90° = ∠OQT
In quad. OPTQ, we have
∠OPT + ∠PTQ + ∠OQT + ∠POQ = 360°
⇒ 90° + ∠PTQ + 90° + 110° = 360°
⇒ ∠PTQ = 70°