如图所示，一个圆穿过三角形 $ABC$ 的顶点。圆在点 $A$ 处的切线与 $BC$ 的延长线相交于点 $D$。点 $C$ 处的切线与 $AD$ 相交于点 $E$。

1. Relate angle EAC to angle ABC using the Alternate Segment Theorem

Since $AD$ is a tangent to the circle at $A$ and $AC$ is a chord, by the Alternate Segment Theorem:
$\angle EAC = \angle ABC$

2. Relate angle ECA to angle ABC using the Alternate Segment Theorem

Similarly, since $EC$ is a tangent to the circle at $C$ and $AC$ is a chord, by the Alternate Segment Theorem:
$\angle ECA = \angle ABC$

3. Use the exterior angle of a triangle to prove the final relationship

In $\triangle EAC$, since $A, E, D$ are collinear, $\angle CED$ is the exterior angle at $E$.
$\angle CED = \angle EAC + \angle ECA$ (Exterior angle of a triangle)
$\angle CED = \angle ABC + \angle ABC$
$\angle CED = 2\angle ABC$ (Shown)

1. Identify a common angle between $\triangle ABD$ and $\triangle CAD$.

In $\triangle ABD$ and $\triangle CAD$,
$\angle ADB = \angle CDA$ (Common angle)

2. Use the alternate segment theorem to find a second pair of equal angles.

$\angle ABD = \angle CAD$ (Alternate segment theorem, since $AD$ is a tangent at $A$ and $AC$ is a chord)

3. Conclude similarity and deduce the required side length relationship.

Therefore, $\triangle ABD$ is similar to $\triangle CAD$ (AA similarity).

Using the ratio of corresponding sides:
$\frac{AD}{CD} = \frac{BD}{AD}$
$AD^2 = BD \times CD$ (Shown)