In the figure (ii) given below, AB is a diameter of a circle with centre O. OD is perpendicular to AB and C is a point on the arc DB. Find ∠BAD and ∠ACD.

In △AOD, ∠AOD = 90°.

OA = OD (Radii of the semi-circle)

∠OAD = ∠ODA (∵ angles opposite equal side are equal.)

We know that sum of angles in a triangle = 180°.

In △OAD,

⇒ ∠AOD + ∠OAD + ∠ODA = 180°
⇒ 90° + ∠OAD + ∠OAD = 180°
⇒ 90° + 2∠OAD = 180°
⇒ 2∠OAD = 180° - 90°
⇒ ∠OAD = $\dfrac{90°}{2}$ = 45°.

From figure,

∠BAD = ∠OAD = 45°.

Arc AD subtends ∠AOD at the centre and ∠ACD at the remaining part of the circle.

∠AOD = 2∠ACD (∵ angle subtended on centre is double the angle subtended at remaining part of the circle.)

⇒ 90° = 2∠ACD
⇒ ∠ACD = $\dfrac{90°}{2}$ = 45°.

Hence, the value of ∠BAD = 45° and ∠ACD = 45°.