图中所示为一个过点 $A$、$B$、$C$ 和 $D$ 的圆。$AC$ 是该圆的直径。直线 $EA$ 是圆的切线，且与直线 $EDC$ 相交于点 $E$。

1. Identify the angle between the tangent and the diameter.

Since $EA$ is a tangent to the circle at $A$ and $AC$ is a diameter, $\angle EAC = 90^\circ$.

2. Identify the angle in a semicircle.

Since $AC$ is a diameter, the angle subtended at the circumference is $90^\circ$, so $\angle ADC = 90^\circ$.

3. Relate the angles in triangle ADE and triangle ADC to show the required result.

In $\triangle AEC$, $\angle AEC + \angle ACE = 180^\circ - 90^\circ = 90^\circ$.
Thus, $\angle AED = 90^\circ - \angle ACE$.
In right-angled $\triangle ADC$, $\angle DAC + \angle ACD = 90^\circ$.
Thus, $\angle DAC = 90^\circ - \angle ACD$.
Since $\angle ACE$ is the same as $\angle ACD$, we have $\angle AED = \angle DAC$.

1. Identify the relevant triangles to prove similar

Consider $\triangle ADE$ and $\triangle CDA$. We need to show they are similar to establish the ratio involving $AD$, $CD$, and $DE$.

2. State the first pair of equal angles

From part (i), we have $\angle AED = \angle DAC$.

3. State the second pair of equal angles

Since $AC$ is a diameter, $\angle ADC = 90^\circ$ (angle in a semicircle).
Since $EDC$ is a straight line, $\angle ADE = 180^\circ - 90^\circ = 90^\circ$.
Therefore, $\angle ADE = \angle CDA = 90^\circ$.

4. Conclude similarity and derive the required equation

By AA similarity, $\triangle ADE$ is similar to $\triangle CDA$.
Therefore, the ratio of corresponding sides is equal:
$\frac{AD}{CD} = \frac{DE}{AD}$
Cross-multiplying gives:
$AD^2 = CD \times DE$