In the given figure, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :

(i) ∠OBD

(ii) ∠AOB

(iii) ∠BED

(i) Given,

AD || BC

∴ OD || BC and BD is transversal.

∠ODB = ∠CBD = 32° [Alternate angles are equal]

In △OBD,

OB = OD [Radius of same circle]

∠OBD = ∠ODB = 32°.

Hence, ∠OBD = 32°.

(ii) Given,

AD || BC

∴ AO || BC and OB is transversal.

∠AOB = ∠OBC [Alternate angles are equal]

From figure,

∠OBC = ∠OBD + ∠DBC = 32° + 32° = 64°.

∴ ∠AOB = 64°.

Hence, ∠AOB = 64°.

(iii) In △OAB,

OA = OB [Radius of same circle]

∠OAB = ∠OBA = x (let)

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ x + x + 64° = 180°

⇒ 2x = 180° - 64°

⇒ 2x = 116°

⇒ x = $\dfrac{116°}{2}$ = 58°

i.e., ∠OAB = 58°.

From figure,

∠DAB = ∠OAB = 58°.

We know that,

Angle in same segment are equal.

∠BED = ∠DAB = 58°.

Hence, ∠BED = 58°.