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: Understanding the Question:
We are given two right-angled triangles, \(\Delta PRQ\) and \(\Delta PSQ\), which share a common hypotenuse \(PQ\).
The line segments \(PR\) and \(QS\) intersect each other at point \(T\).
We need to prove that \(ST \times TQ = PT \times TR\).

Step 2: Key Formula or Approach:
1. Since \(\angle PRQ = 90^\circ\) and \(\angle PSQ = 90^\circ\) are subtended by the same segment \(PQ\), the points \(P\), \(S\), \(R\), and \(Q\) must lie on a circle where \(PQ\) is the diameter.
2. This makes \(PQRS\) a cyclic quadrilateral.
3. We can establish similarity between \(\Delta PTS\) and \(\Delta QTR\) using the angle properties of circles (angles subtended by the same arc are equal).

Step 3: Detailed Explanation:
1. Consider the circle with diameter \(PQ\). Since \(\angle PRQ = 90^\circ\) and \(\angle PSQ = 90^\circ\), the points \(P\), \(S\), \(R\), and \(Q\) lie on the circumference of this circle.
2. Therefore, \(PQRS\) is a concyclic set of points.
3. Now, compare triangles \(\Delta PTS\) and \(\Delta QTR\):
- \(\angle PTS = \angle QTR\) (Vertically opposite angles are equal)
- \(\angle SPT = \angle RQT\) (Angles subtended by the same arc \(SR\) at the circumference are equal)
4. By the Angle-Angle (AA) Similarity Criterion:
\[ \Delta PTS \sim \Delta QTR \] 5. Since corresponding sides of similar triangles are proportional:
\[ \frac{PT}{QT} = \frac{ST}{RT} \] 6. Cross-multiplying the ratios yields:
\[ PT \times RT = QT \times ST \] \[ ST \times TQ = PT \times TR \] 7. Hence, the relation is successfully proved.

Step 4: Final Answer:
The required statement \(ST \times TQ = PT \times TR\) is proved.