在图中（未按比例绘制），点 $B$、$Q$、$R$ 和 $S$ 均位于圆周上。$ABC$ 是过点 $B$ 的圆的切线。弦 $BR$ 和 $QS$ 交于点 $P$。

1. Use the given isosceles triangle property to relate angles.

Since $BS = SR$, $\triangle BSR$ is an isosceles triangle.
Therefore, $\angle SBR = \angle SRB$.

2. Apply the alternate segment theorem to relate an angle in the alternate segment to the angle between the tangent and chord.

By the alternate segment theorem, the angle between the tangent $ABC$ and the chord $BS$ is equal to the angle in the alternate segment.
Thus, $\angle ABS = \angle SRB$.

3. Conclude that BS bisects angle ABR.

Since $\angle SBR = \angle SRB$ and $\angle ABS = \angle SRB$, it follows that $\angle ABS = \angle SBR$.
Therefore, $BS$ bisects $\angle ABR$.

1. Express the angles of $\triangle BSR$ in terms of $\angle BSQ$.

Let $\angle BSQ = x$.
Since $QS$ bisects $\angle BSR$, $\angle BSR = 2x$.
In $\triangle BSR$, $BS = SR$ (given), so it is an isosceles triangle.
$\angle SRB = \angle SBR = \frac{180^\circ - 2x}{2} = 90^\circ - x$.

2. Find $\angle SQB$ using angles in the same segment.

Angles subtended by the same arc $SB$ at the circumference are equal:
$\angle SQB = \angle SRB = 90^\circ - x$.

3. Calculate $\angle SBQ$ and conclude that $SQ$ is the diameter.

In $\triangle BSQ$, the sum of angles is $180^\circ$.
$\angle SBQ = 180^\circ - \angle BSQ - \angle SQB$
$\angle SBQ = 180^\circ - x - (90^\circ - x) = 90^\circ$.
Since $\angle SBQ = 90^\circ$, the chord $SQ$ subtends a right angle at the circumference.
Therefore, $SQ$ is the diameter of the circle (converse of angle in a semicircle).