Q7 of 31 Page 209

P is any point on diameter of a circle with center O. A perpendicular drawn on diameter at the point O intersects the circle at the point Q. Extended QP intersects the circle at the point R. A tangent drawn at the point R intersects extended OP at the point S Let us prove that SP = SR

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



In Δ QPO


∠QOP = 90°


By angle sum property


∠QOP + ∠QPO + ∠OQP = 180°


∠QPO = 90° - ∠OQP


In Δ QOR


As OQ = OR ∵ radius of same circle


∠OQP = ∠ORP


As RS is tangent


∠ORS = 90°


In Δ SPR


∠SPR = ∠OPQ ∵ (Vertically opposite angles)


∠SPR = 90° - ∠OQP


∠SRP = ∠ORS - ∠ORP


= 90° - ∠OQP


Hence;


∠SPR = ∠SRP


By isosceles triangle property


∴ SR = SP


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