In the given figure, $PQ$ is tangent to a circle centred at $O$ and $\angle BAQ = 30°$; show that $BP = BQ$.

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Question

In the given figure, $PQ$ is tangent to a circle centred at $O$ and $\angle BAQ = 30°$; show that $BP = BQ$.

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Accepted Answer

Step 1: Circle with tangent PQ, points A, B, O, P, Q and angles 1-6 marked
Step 2: Join $OQ$. $OQ = OA$ (radii) $\Rightarrow \angle 2 = 30°$
Step 3: $\angle 3 = 90° - 30° = 60°$
Step 4: $\angle 4 = 90° - 60° = 30°$
Step 5: $\angle 6 = \angle 1 + \angle 2 = 60°$. Hence $\angle 5 = 90° - 60° = 30° = \angle 4$
Step 6: $	herefore BP = BQ$

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