Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that :

$∠ P A Q = 2 ∠ O P Q$

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Solution

Let $∠ O P Q = x N o w w e k n o w t h a t, ∠ O P A = 90 (R a d i u s i s a l w a y s p e r p e n d i c u l a r t o t a n g e n t) H e n c e, ∠ Q P A = 90 - x ∠ Q P A = ∠ P Q A = 90 - x . (A P = A Q t a n g e n t s f r o m s a m e e x t e r n a l p t s) N o w b y a n g l e s u m p r o p e r t y i n ∠ A P Q ∠ Q P A + ∠ P Q A + ∠ Q A P = 180 S u b s t i t u t i n g t h e v a l u e s 90 - x + 90 - x + ∠ Q A P = 180 ∠ Q A P - 2 x = 0 ∠ Q A P = 2 x O r w e c a n s a y t h a t, ∠ Q A P = 2 ∠ O P Q$

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