In the figure (ii) given below, AB is a diameter of the semicircle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also prove that OE is parallel to BD.

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

In cyclic quadrilateral BCDE,

⇒ ∠BCD + ∠BED = 180°
⇒ 140° + ∠BED = 180°
⇒ ∠BED = 180° - 140° = 40°.

∠AEB = 90°. (∵ angle in semicircle = 90°.)

From figure,

∠AED = ∠AEB + ∠BED = 90° + 40° = 130°.

In cyclic quadrilateral AEDB,

⇒ ∠AED + ∠DBA = 180°
⇒ 130° + ∠DBA = 180°
⇒ ∠DBA = 180° - 130° = 50°.

Given chord AE = ED

∴ ∠DBE = ∠EBA

From figure,

⇒ ∠DBA = ∠DBE + ∠EBA
⇒ 50 = ∠DBE + ∠DBE
⇒ 2∠DBE = 50°
⇒ ∠DBE = $\dfrac{50°}{2}$ = 25°.

or, ∠EBD = 25°.

In △OEB, OE = OB (Radii of the same circle.)

∠OEB = ∠EBO = ∠DBE

But these are alternate angles.

∴ OE || BD.

Hence, the value of ∠AED = 130° and ∠EBD = 25°.