In the diagram, BDK is a straight line and AB = AD. Angle ABD = 36°, angle CBD = 49° and angle CDK = 121°. Explain why it is possible to draw a circle that passes through the points A, B, C and D. Give reasons for each step of your working. Answer ...

1. Find $\angle BAD$ using the isosceles triangle $ABD$.

Since $AB=AD$, $\triangle ABD$ is isosceles, so
$\angle ADB=\angle ABD=36^\circ$.

Hence
$\angle BAD=180^\circ-36^\circ-36^\circ=108^\circ.$

2. Find $\angle BCD$ in triangle $BCD$.

Since $BDK$ is a straight line,
$\angle CDB=180^\circ-121^\circ=59^\circ.$

In $\triangle BCD$,
$\angle BCD=180^\circ-49^\circ-59^\circ=72^\circ.$

3. Use the cyclic quadrilateral condition to justify that a circle can pass through all four points.

Now
$\angle BAD+\angle BCD=108^\circ+72^\circ=180^\circ.$

Opposite angles of quadrilateral $ABCD$ are supplementary. Therefore, by the converse of the cyclic quadrilateral angle property, $A,B,C,D$ lie on the same circle.