如图，BDK 是一条直线，且 AB = AD。角 ABD = 36°，角 CBD = 49°，角 CDK = 121°。 解释为什么可以画一个圆通过点 A、B、C 和 D。为你的每一步推导给出理由。 答案 ...

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.