In the diagram, A, B and C lie on a circle where O is the centre and AC is a diameter. The tangents to the circle at B and C meet at Q. The line BO extended meets the line QC at P.

1. Identify the relationship between the tangent and the radius at the point of contact.

Since $QC$ is a tangent to the circle at $C$ and $OC$ is a radius, $\angle OCQ = 90^\circ$.

2. Express angle OCB in terms of angle BCQ.

$\angle OCB = \angle OCQ - \angle BCQ = 90^\circ - \angle BCQ$.

3. Use the property of isosceles triangle OBC to find angle BOC.

In $\triangle OBC$, $OB = OC$ (radii of the same circle), so $\triangle OBC$ is an isosceles triangle.
Therefore, $\angle OBC = \angle OCB = 90^\circ - \angle BCQ$.
The sum of angles in $\triangle OBC$ is $180^\circ$, so:
$\angle BOC = 180^\circ - (\angle OBC + \angle OCB)$
$\angle BOC = 180^\circ - 2(90^\circ - \angle BCQ)$
$\angle BOC = 180^\circ - 180^\circ + 2\angle BCQ$
$\angle BOC = 2\angle BCQ$ (Shown)

1. Find the right angle in each triangle using the tangent–radius property

$OC$ is a radius and $QC$ (which contains $P$) is the tangent at $C$, so $OC \perp QC$.
$\angle PCO = 90^\circ$
Similarly, $OB$ is a radius and $QB$ is the tangent at $B$, and since $B$, $O$, $P$ are collinear, $OB \perp QB$ gives
$\angle PBQ = 90^\circ$
Hence $\angle PCO = \angle PBQ = 90^\circ$.

2. Identify the common angle and conclude AA similarity

$\angle OPC$ (in $\triangle POC$) and $\angle QPB$ (in $\triangle PQB$) are the same angle since $P$ lies on both line $BOP$ and line $QCP$:
$\angle OPC = \angle QPB \quad (\text{common angle})$
With two pairs of equal angles,
$\triangle POC \sim \triangle PQB \quad (\text{AA})$