In the diagram, A, B, C and P lie on a circle. The line PQ is the tangent to the circle at P. The tangent meets CA produced at Q. The lines AC and PB intersect at D. AP bisects angle 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:
$AB^2 = AD \times AC$