Q3 of 31 Page 209

I drew a circle having PR as a diameter. I draw a tangent at tangent at the point P and a point S is taken on the tangent of the circle in such a way that PR = PS. If RS intersects the circle at the point T. Let us prove that ST = RT = PT.

Theory.


⇒ Angle sum property of triangle is 180°


⇒ if 2 sides of triangle are equal then their corresponding angles will also be equal


Solution.



In Δ PRS


PS = PR


∴ ∠PSR = ∠PRS


∠RPS = 90° ∵ (Radius of circle from point of contact of tangent is 90° )


∠PSR + ∠PRS + ∠RPS = 180°


2∠PSR = 180° - 90°


∠PSR = 45°


∠PSR = ∠PRS = 45°


In Δ PRT


∠PTR = 90° ∵ (3rd point of triangle on circumference of semicircle is always 90° )


∠PRT = ∠PRS = 45°


∠TPR + ∠PRT + ∠PTR = 180°


∠TPR = 180° - 135°


= 45°


∠PRT = ∠TPR


RT = TP ∵ (isosceles triangle property)………1


In Δ PTS


∠RPS = ∠TPS + ∠TPR = 90°


∠TPS + 45° = 90°


∠TPS = 45°


∠PST = ∠PSR = 45° ∵ (proved above)


∠PST = ∠TPS


PT = ST ∵ (isosceles triangle property)………2


Joining 1 and 2


We get ;


PT = ST = RT


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 2

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