In the diagram, $AP$ is a tangent to the circle passing through $A$, $B$ and $C$.
$AC$ is the diameter of the circle and $PBC$ is a straight line.
The bisector of angle $APC$ meets $AB$ at $H$ and $AC$ at $K$.

1. Find the relationship between angle PBA and angle ABC.

Since $AC$ is the diameter of the circle, $\angle ABC = 90^\circ$ (angle in a semicircle).
Since $PBC$ is a straight line, $\angle PBA = 180^\circ - 90^\circ = 90^\circ$ (angles on a straight line).
Therefore, $\angle PBA = \angle ABC$.

2. Identify a second pair of equal angles using the tangent.

Since $AP$ is a tangent to the circle at $A$ and $AB$ is a chord, $\angle PAB = \angle ACB$ (alternate segment theorem).

3. Conclude the similarity of the triangles.

In $\triangle BAP$ and $\triangle BCA$:

1. $\angle PBA = \angle ABC = 90^\circ$
2. $\angle PAB = \angle BCA$
   By AA similarity, $\triangle BAP \sim \triangle BCA$.

1. Define variables for the bisected angle and find $\angle AHK$.

Let $\angle APK = \angle CPK = x$.
Since $AC$ is a diameter, $\angle ABC = 90^\circ$ (angle in a semicircle).
As $PBC$ is a straight line, $\angle PBA = 180^\circ - 90^\circ = 90^\circ$.
In $\triangle PBH$, $\angle PHB = 180^\circ - 90^\circ - x = 90^\circ - x$.
$\angle AHK = \angle PHB = 90^\circ - x$ (vertically opposite angles).

2. Find $\angle AKH$ and conclude the proof.

Since $AP$ is a tangent and $AC$ is a diameter, $\angle CAP = 90^\circ$ (tangent $\perp$ radius).
In $\triangle APC$, $\angle ACP = 180^\circ - 90^\circ - 2x = 90^\circ - 2x$.
Considering $\triangle PKC$, the exterior angle $\angle AKH = \angle CPK + \angle ACP$.
$\angle AKH = x + (90^\circ - 2x) = 90^\circ - x$.
Since $\angle AHK = \angle AKH = 90^\circ - x$, $\triangle AHK$ is an isosceles triangle.
Therefore, $AH = AK$.

1. State the ratio of corresponding sides from the given similar triangles.

Since $\triangle BHP$ is similar to $\triangle AKP$, the ratio of their corresponding sides is equal:
$\frac{BH}{AK} = \frac{HP}{KP}$

2. Cross-multiply and substitute the result from part (ii) to complete the proof.

Cross-multiplying gives:
$BH \times KP = AK \times HP$
From part (ii), we know that $AK = AH$. Substituting this into the equation:
$BH \times KP = AH \times HP$
Rearranging yields:
$AH \times HP = KP \times BH$