Given : $\angle PSQ = 90^\circ$ and $\angle PRQ = 90^\circ$; $PR$ and $QS$ intersect at $T$.
To Prove : $ST \times TQ = PT \times TR$.

Step 1: In $\triangle PTS$ and $\triangle QTR$ :
$$\angle PST = \angle QRT = 90^\circ \quad \text{(given right angles)}$$
$$\angle PTS = \angle QTR \quad \text{(vertically opposite angles)}$$
Step 2: By AA similarity criterion,
$$\triangle PTS \sim \triangle QTR$$
Step 3: Corresponding sides of similar triangles are proportional :
$$\frac{PT}{QT} = \frac{ST}{RT}$$
Step 4: Cross-multiplying :
$$PT \times RT = QT \times ST$$
$$\Rightarrow ST \times TQ = PT \times TR$$
Hence proved.