In the diagram, $PQS$ lies on a circle and $PCBO$ lies on another circle. PQ is a common chord of the two circles. BPS is a straight line. QC is a tangent to the circle PQS at Q. QC and BP intersect at M.

1. Relate the angle between the tangent and the chord to the angle in the alternate segment.

$\angle PQC = \angle PSQ$ (Alternate segment theorem)

2. Relate angles subtended by the same arc in the second circle.

$\angle PQC = \angle PBC$ (Angles in the same segment, subtended by arc $PC$)

3. Conclude that the lines are parallel using the equality of alternate angles.

Therefore, $\angle PSQ = \angle PBC$.
Since $BPS$ is a straight line, $\angle PSQ$ and $\angle PBC$ are alternate angles.
Because the alternate angles are equal, $BC \parallel QS$.

1. Identify the common angle between the two triangles.

In $\triangle PQM$ and $\triangle QSM$,
$\angle PMQ = \angle QMS$ (Common angle)

2. Use the alternate segment theorem to find a second pair of equal angles.

Since $QC$ is a tangent to the circle $PQS$ at $Q$ and $PQ$ is a chord, the angle between the tangent $MQ$ and chord $PQ$ is equal to the angle in the alternate segment.
$\angle MQP = \angle PSQ$ (Alternate segment theorem)
Since $S, P, M$ are collinear, $\angle PSQ = \angle MSQ$.
Therefore, $\angle MQP = \angle MSQ$.

3. Conclude that the triangles are similar.

Since two pairs of corresponding angles are equal, $\triangle PQM$ is similar to $\triangle QSM$ (AA similarity).

1. State the ratio of corresponding sides from the similar triangles proven in part (b).

Since $\triangle PQM$ is similar to $\triangle QSM$, the ratio of their corresponding sides is equal:
$\frac{QM}{SM} = \frac{PM}{QM}$

2. Rearrange the ratio to show the required equation.

Cross-multiplying gives:
$QM \times QM = PM \times SM$
$QM^2 = PM \times SM$