ABCD is a circle with center O. $\angle AOB = 80^{\circ}$ and $AB = BC$. We need to calculate (i) $\angle ACB$, (ii) $\angle ADC$, and (iii) $\angle OAC$.

1. Problem Description

ABCD is a circle with center O. $\angle AOB = 80^{\circ}$ and $AB = BC$.

We need to calculate (i) $\angle ACB$, (ii) $\angle ADC$, and (iii) $\angle OAC$.

2. Solution Steps

(i) To find $\angle ACB$:

Since $AB = BC$, triangle $ABC$ is an isosceles triangle. Let $\angle BAC = \angle BCA = x$. The angle at the center is twice the angle at the circumference. Therefore, $\angle AOB = 2 \angle ACB$. So, $80^{\circ} = 2 \angle ACB$.

$\angle ACB = \frac{80}{2}$

$\angle ACB = 40^{\circ}$

(ii) To find $\angle ADC$:

The sum of opposite angles in a cyclic quadrilateral is $180^{\circ}$.

Therefore, $\angle ADC + \angle ABC = 180^{\circ}$.

Since the sum of angles in triangle AOB is 180, $\angle OAB = \angle OBA = (180 - 80)/2 = 50^{\circ}$.

Since AB = BC, let angle $BAC$ = angle $BCA = x$. Also, $ABC$ is a triangle. Then, the angle $ABC$ is given by

$\angle ABC = \angle ABO + \angle OBC$.

Since the angle subtended by the arc AC at the center is $\angle AOC$, $\angle AOC = 360 - 80 - (360 -80 -2*y) = 360 -80 = 280$. where y is the angle BOC.

$2x + \angle ABC = 180^{\circ}$.

Also, $\angle AOC = 2 \angle ADC$.

Angle subtended at center is twice angle at circumference.

Since $AB=BC$, angle $BOC = angle AOB = 80^{\circ}$.

So, angle $ABC = \angle ABO + \angle CBO = 50 + 50 = 100$. This is incorrect.

$\angle ABC + \angle ADC = 180^{\circ}$.

$\angle AOB = 80^{\circ}$ and since $AB = BC$, $\angle BOC = \angle AOB = 80^{\circ}$.

$\angle AOC = \angle AOB + \angle BOC = 80 + 80 = 160^{\circ}$.

$\angle ADC = \frac{1}{2} \angle AOC = \frac{160}{2} = 80^{\circ}$.

(iii) To find $\angle OAC$:

Since $OA$ and $OC$ are radii of the circle, triangle $OAC$ is an isosceles triangle with $OA = OC$.

Thus $\angle OAC = \angle OCA$.

Since $\angle AOC = 160^{\circ}$, then

$\angle OAC = \angle OCA = \frac{180 - 160}{2} = \frac{20}{2} = 10^{\circ}$.

3. Final Answer

(i) $\angle ACB = 40^{\circ}$

(ii) $\angle ADC = 80^{\circ}$

(iii) $\angle OAC = 10^{\circ}$