Q. 84.0( 2 Votes )

# Rumela drew a circle with centre with centre O of which QR is a chord. Two tangents drawn at the points Q and R intersect each other at the point P. If QM is a diameter, let us prove that ∠QPR = 2 ∠RQM.

Answer :

Formula used.

⇒ Isosceles triangle property

If 2 angles of triangle are equal then their corresponding sides are also equal

⇒ Perpendicular drawn through tangent pass through centre

Solution

Join OR

As QP is tangent at point Q and RP tangent to point R

Hence;

∠OQP = ∠ORP = 90°

In Δ OQR

As OQ = OR

By isosceles triangle property

∠OQR = ∠ORQ

In Δ PQR

By angle sum property

∠P + ∠PQR + ∠PRQ = 180°

∠P + [90° - ∠OQR] + [90° - ∠ORQ] = 180°

∠P + ∠OQR + ∠ORQ = 180° - 180°

∠P = ∠OQR + ∠ORQ

∠P = 2∠OQR

∠P = 2∠MQR

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