Ex 10.2

Chapter 10 Class 10 Circles
Serial order wise

### 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

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.