Q. 34.5( 4 Votes )

# 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.

Answer :

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° ∵ (3^{rd} 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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