web analytics

Ex 10.2, 9 - In fig, XY and X′Y′ are two parallel tangents - Theorem 10.2: Equal tangents from external point (proof type)

  1. Chapter 10 Class 10 Circles
  2. Serial order wise
Ask Download

Transcript

Ex 10.2,9 In figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′at B. Prove that ∠AOB = 90°. Given : XY is a tangent at point P and X’Y’ is a tangent at point Q And XY ∥ X’Y’ AB is a tangent at point C To prove: ∠ AOB = 90° Proof: Join OC For tangent AB & Radius OC OC ⊥ AB So, ∠ ACO = ∠ BCO = 90° In Δ AOP & Δ AOC OP = OC AP = AC OA = OA ∴ Δ AOC ≅ Δ AOP So, ∠ AOP= ∠ AOC Now In Δ BOC & Δ BOQ OC = OQ BC = BQ OB = OB ∴ Δ BOC ≅ Δ BOQ So, ∠ BOC = ∠ BOQ For line PQ ∠ AOP + ∠ AOC + ∠ BOC + ∠ BOQ = 180° ∠ AOC + ∠ AOC + ∠ BOC + ∠ BOC = 180° 2∠ AOC + 2∠ BOC = 180° 2(∠ AOC + ∠ BOC) = 180° ∠ AOC + ∠ BOC = (180°)/2 ∠ AOC + ∠ BOC = 90° ∠ AOB = 90° Hence proved

About the Author

CA Maninder Singh's photo - Expert in Practical Accounts, Taxation and Efiling
CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
Jail