Hi!

Here is the answer to your question.

Given: A circle with centre O. PQ and PR are tangents drawn from an external point P to the circle.

To prove: ∠QPR = 2∠OQR

Proof:

∠OQP = ∠ORP = 90° (Radius is perpendicular to the tangent at point of contact)

In quadrilateral OQPR

∠QPR + ∠OQP + ∠QOR + ∠ORP = 360°

∴∠OPR + 90° + ∠QOR + 90° = 360°

⇒ ∠OPR + ∠QOR = 180° ... (1)

In ∆OQR

OQ = OR (Radius of the circle)

⇒ ∠ORQ = ∠OQR (Equal sides have equal angles opposite to them)

∠QOR + ∠OQR + ∠ORQ = 180° (angle sum property)

⇒ ∠QOR + 2∠OQR = 180° ... (2)

From (1) and (2), we get

∠QPR + ∠QOR = ∠QOR + 2∠OQR

⇒ ∠QPR = 2∠OQR

Cheers!

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