Mathematics
Tangents AP and AQ are drawn to a circle, with center O, from an exterior point A. Prove that :
∠PAQ = 2∠OPQ
Answer
We know that,
The tangent at any point of a circle and the radius through this point are perpendicular to each other.
In quadrilateral OPAQ,
∠OPA = ∠OQA = 90°.
∠OPA + ∠OQA + ∠POQ + ∠PAQ = 360° [∵ Sum of angles in quadrilateral = 360°]
90° + 90° + ∠POQ + ∠PAQ = 360°
∠POQ + ∠PAQ = 360° - 180°
∠POQ + ∠PAQ = 180° ……….(1)
In △OPQ,
OP = OQ [Radius of same circle]
∴ ∠OPQ = ∠OQP [Angles opposite to equal sides are equal]
By angle sum property of triangle,
⇒ ∠OPQ + ∠OQP + ∠POQ = 180°
⇒ ∠OPQ + ∠OPQ + ∠POQ = 180°
⇒ 2∠OPQ + ∠POQ = 180° ………(2)
From (1) and (2) we get,
∠POQ + ∠PAQ = 2∠OPQ + ∠POQ
⇒ ∠PAQ = 2∠OPQ.
Hence, proved that ∠PAQ = 2∠OPQ.
Related Questions
From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that:
(i) ∠AOP = ∠BOP
(ii) OP is the ⊥ bisector of chord AB.
ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with center O, has been inscribed inside the triangle. Calculate the value of x, the radius of the inscribed circle.
In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that :
(i) tangent at point P bisects AB.
(ii) angle APB = 90°.
In a triangle ABC, the incircle (center O) touches BC, CA and AB at points P, Q and R respectively. Calculate :
(i) ∠QOR
(ii) ∠QPR;
given that ∠A = 60°.