Question

Two right triangles PRQ and PSQ are drawn on the same hypotenuse PQ. If PR and QS intersect at T, prove that ST \(\times\) TQ = PT \(\times\) TR.

Hint:

Whenever you see right angles sharing a common hypotenuse, think of them as lying on a circumscribed circle.
The intersecting chords theorem then states that for any two chords intersecting at \(T\), the products of their segments are equal: \(PT \times TR = ST \times TQ\).

Answer 1

Step 1: Establish that all four points lie on one circle.
Since $\angle PRQ = 90^{\circ}$ and $\angle PSQ = 90^{\circ}$ are both angles in a semicircle on the same segment $PQ$, points $P$, $R$, $Q$, and $S$ all lie on a circle that has $PQ$ as its diameter.

Step 2: Identify the two chords through the intersection point.
Inside this circle, $PR$ and $QS$ are two chords, and we are told they cross each other at the point $T$.

Step 3: Apply the intersecting chords theorem directly, instead of proving triangles similar.
The intersecting chords theorem states that when two chords of a circle cross at an interior point, the product of the two parts of one chord equals the product of the two parts of the other chord. For chords $PR$ and $QS$ meeting at $T$:
\[ PT \times TR = QT \times TS \]

Step 4: Rewrite in the form asked for in the question.
Since $QT \times TS$ is the same as $ST \times TQ$, we get:
\[ ST \times TQ = PT \times TR \]

Final Answer:
The required relation $ST \times TQ = PT \times TR$ follows directly from the intersecting chords theorem, since $P$, $S$, $R$, $Q$ are concyclic.
\[ \boxed{ST \times TQ = PT \times TR} \]