In the given figure, PQ is tangent to a circle at O and , show that BP = BQ.

Question

In the given figure, PQ is tangent to a circle at O and $∠ B A Q = 30 °$ , show that BP = BQ.

Smart Achievers · Accepted Answer

$∠ B Q P = ∠ B A Q (∠ s in alternate segment are equal)$

$\Rightarrow ∠ B Q P = 30 ° ...(i) (∵ ∠ B A Q = 30 ° given)$

As AB is a diameter, AQB is a Semicircle.

$∠ A Q B = 90 ° (angle in semicircle = 90 °)$

From figure, $∠ A Q P = ∠ A Q B + ∠ B Q P$

$\Rightarrow ∠ A Q P = 90 ° + 30 ° = 120 °$

In $∆ A Q P,$ $∠ Q P A + ∠ B A Q + ∠ A Q P = 180 °$

$\Rightarrow ∠ Q P A + 30 ° + 120 ° = 180 °$

$\Rightarrow ∠ Q P A = 180 ° - (30 ° + 120 °)$

$\Rightarrow ∠ Q P A = 30 ° ...(ii)$

From (i) and (ii) we get

$∠ B Q P = ∠ Q P B = 30 °$

Therefore, QB = BP

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