Question

Two tangents ap and aq are drawn to a circle with centre o from an external pt a prove paq = 2 opq

Answer 1

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Answer 2

Tangent drawn to a Circle is ⊥ to its Radius

Step-by-step explanation:

In Quadrilateral OPAQ,

$\angle OPA = \angle OQA = 90$°

(As OP ⊥ PA and OQ ⊥ QA)

So,  

$\angle POQ + \angle PAQ$ + 90° + 90° = 360 °

⇒ $\angle POQ + \angle PAQ = 360 - 180$

⇒ $\angle POQ + \angle PAQ = 180$° ""(1)

In ΔOPQ,  

OP = OQ (Radii of same circle)

So, $\angle OPQ = \angle OQP$

But  

$\angle POQ + \angle OPQ + \angle OQP = 180$°

=  $\angle POQ + 2 \angle OPQ = 180$° """(2)

From (1) & (2) we get-

$\angle POQ + \angle PAQ = \angle POQ + 2 \angle OPQ$

⇒ $\angle PAQ = 2 \angle OPQ$