Question

In the adjoining figure, O is the centre of a circle. Chord CD is parallel to diameter AB. If $\angle ABC={25}^{\circ }$, calculate $\angle CED$.

Answer 1

∠BCD = ∠ABC = 25° (Alternate angles)
Join CO and DO.
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by an arc at any point on the circumference.
Thus, ∠BOD = 2∠BCD
⇒∠BOD = 2 × 25° = 50°
Similarly, ∠AOC = 2∠ABC
⇒ ∠AOC = 2 × 25° = 50°
AB is a straight line passing through the centre.
i.e., ∠AOC + ∠COD + ∠BOD = 180°
⇒ 50° + ∠COD + 50° = 180°
⇒ ∠COD = (180° – 100°) = 80°
⇒∠CED=$\frac{1}{2}$​​​​​​​∠COD
⇒∠CED=($\frac{1}{2}$×80°)=40°
∴ ∠CED = 40°