在图中，A、B、C 和 P 共圆。直线 PQ 是圆在点 P 处的切线。该切线与 CA 的延长线相交于点 Q。直线 AC 和 PB 相交于点 D。AP 平分角 QPB。

1. Use the angle bisector property to relate angles at P.

Since $AP$ bisects $\angle QPB$, we have:
$\angle QPA = \angle APB$

2. Apply the alternate segment theorem.

By the alternate segment theorem, the angle between the tangent $PQ$ and chord $PA$ is equal to the angle in the alternate segment:
$\angle QPA = \angle ABP$

3. Conclude that the triangle is isosceles.

Therefore, $\angle ABP = \angle APB$.
Since two angles in triangle $ABP$ are equal, triangle $ABP$ is isosceles (with $AB = AP$).

1. Identify a common angle between the two triangles.

In $\triangle ACP$ and $\triangle APD$, $\angle CAP$ is common to both triangles (i.e., $\angle CAP = \angle DAP$).

2. Use the alternate segment theorem and angle bisector property to find another pair of equal angles.

By the alternate segment theorem, $\angle ACP = \angle QPA$.
Since $AP$ bisects $\angle QPB$, $\angle QPA = \angle APB$.
Therefore, $\angle ACP = \angle APB$ (which is $\angle APD$).

3. Conclude similarity using the AA criterion.

Since two pairs of corresponding angles are equal ($\angle CAP = \angle DAP$ and $\angle ACP = \angle APD$), $\triangle ACP$ is similar to $\triangle APD$ (AA similarity).

1. Use the similarity of triangles ACP and APD to establish a relationship between their sides.

Since $\triangle ACP$ is similar to $\triangle APD$ (from part (b)(i)), the ratio of their corresponding sides is equal:
$\frac{AC}{AP} = \frac{AP}{AD}$
Cross-multiplying gives:
$AP^2 = AD \times AC$

2. Substitute the result from part (a) to complete the proof.

From part (a), $\triangle ABP$ is isosceles with $AB = AP$.
Substituting $AP$ with $AB$ into the equation yields:
$AB^2 = AD \times AC$