In the adjoining figure, O is the center and AOE is the diameter of the semicircle ABCDE. If AB = BC and ∠AEB = 50°, find :

(i) ∠CBE

(ii) ∠CDE

(iii) ∠AOB.

Prove that OB is parallel to EC.

(i) AECB is a cyclic quadrilateral as all of its vertices lie on the circumference of the circle.

From figure,

∠ABE = 90° (∵ angle in semicircle is 90°.)

We know that sum of opposite angles of a cyclic quadrilateral = 180°.

⇒ ∠AEC + ∠ABC = 180°
⇒ ∠AEC + ∠ABE + ∠CBE = 180°
⇒ 50° + 90° + ∠CBE = 180°
⇒ ∠CBE + 140° = 180°
⇒ ∠CBE = 180° - 140° = 40°.

Hence, the value of ∠CBE = 40°.

(ii) BEDC is a cyclic quadrilateral as all of its vertices lie on the circumference of the circle.

We know that sum of opposite angles of a cyclic quadrilateral = 180°.

⇒ ∠CBE + ∠CDE = 180°
⇒ 40° + ∠CDE = 180°
⇒ ∠CDE = 180° - 40° = 140°.

Hence, the value of ∠CDE = 140°.

(iii) Given,

AB = BC

∴ ∠AEB = ∠BEC = $\dfrac{1}{2}$∠AEC = $\dfrac{1}{2} \times 50° = 25°.$ (∵ equal chords subtend equal angle at circumference.)

In △OBE,

OB = OE = radius of the same circle

∴ ∠OBE = ∠OEB = 25°.

In triangle exterior angle is equal to the sum of opposite two interior angles.

∠AOB = ∠OBE + ∠OEB = 25° + 25° = 50°.

Hence, the value of ∠AOB = 50°.

∠AOB = ∠OEC (∵ both are equal to 50°)

Since these angles are corresponding angles and are equal which is property of parallel lines.

Hence proved that OB || EC.