Q8 of 31 Page 209

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.

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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