Question

in the adjoining figure it is given that OA parallel to EC and OB parallel to ED prove that angle ABC= angle CED.

Please help me guys.​

Answer 1

Answer:

1.∠BCD = ∠ABC (alternate ∠s in AB//CD) . . . so ∠BCD = 35°

2.∠OCB = ∠OBC (base ∠s of isosceles ∆OCB) . . . so ∠OCB = 35°

3.∠OCD = ∠OBC + ∠BCD = 35° + 35° . . . so ∠OCD= 70°

4.∠ODC = ∠OCD (base ∠s of isosceles ∆OCD) . . . so ∠ODC = 70°

5.∠COD = 180° - ∠OCD - ∠ODC (sum of interior ∠s of ∆OCD) = 180° - 70° - 70° = 40° . . . so ∠COD = 40°

6.Finally, ∠CED = ∠COD2 (∠ at circumference = half ∠ at centre)