In the diagram, O is the centre of the circle and ABCD is a cyclic quadrilateral.
Angle $AOB = \angle ADC = 70^\circ$.
Explain why triangle $ABC$ is an isosceles triangle.
Give reasons for each step of your working.

1. Find $\angle ACB$ using the angle at the centre theorem.

$\angle AOB=70^\circ$.

The angle at the centre is twice the angle at the circumference standing on the same arc $AB$.

Hence
$\angle ACB=\frac{1}{2}\times 70^\circ=35^\circ.$

2. Find $\angle ABC$ using the property of opposite angles in a cyclic quadrilateral.

Since $ABCD$ is a cyclic quadrilateral, opposite angles are supplementary.

$\angle ABC+\angle ADC=180^\circ$
$\angle ABC+70^\circ=180^\circ$
$\angle ABC=110^\circ.$

3. Show that triangle $ABC$ has two equal angles, then conclude it is isosceles.

In triangle $ABC$,
$\angle BAC=180^\circ-\angle ABC-\angle ACB$
$=180^\circ-110^\circ-35^\circ$
$=35^\circ.$

So
$\angle BAC=\angle ACB=35^\circ.$

Therefore the opposite sides are equal:
$BC=AB.$
Hence triangle $ABC$ is isosceles.