NCERT Solutions
Class 10 - Mathematics - Chapter 6: TRIANGLES - Exercise 6.3

Question 4

Given: In Fig. 6.36,

• \(\dfrac{QR}{QS} = \dfrac{QT}{PR}\)
• \(\angle 1 = \angle 2\)

To prove: \(\triangle PQS \sim \triangle TQR\).

Step 1: Use the angle condition in \(\triangle PQR\)

Angles \(\angle 1\) and \(\angle 2\) are the base angles of \(\triangle PQR\):

\[\angle 1 = \angle PQR, \quad \angle 2 = \angle PRQ\]

Given \(\angle 1 = \angle 2\), so

\[\angle PQR = \angle PRQ.\]

In a triangle, sides opposite equal angles are equal, therefore

\[PQ = PR. \tag{1}\]

Step 2: Rewrite the given side ratio using (1)

Given:

\[\frac{QR}{QS} = \frac{QT}{PR}.\]

Using \(PR = PQ\) from (1):

\[\frac{QR}{QS} = \frac{QT}{PQ}. \tag{2}\]

Step 3: Compare sides around the common angle at \(Q\)

Consider \(\triangle PQS\) and \(\triangle TQR\).

• In \(\triangle PQS\), the sides including \(\angle Q\) are \(QP\) and \(QS\).
• In \(\triangle TQR\), the sides including \(\angle Q\) are \(QT\) and \(QR\).

From (2):

\[\frac{QR}{QS} = \frac{QT}{QP} \quad \Longrightarrow \quad \frac{QS}{QR} = \frac{QP}{QT}. \tag{3}\]

Also, \(\angle PQS\) and \(\angle TQR\) are the same angle at \(Q\):

\[\angle PQS = \angle TQR = \angle Q. \tag{4}\]

Step 4: Apply SAS similarity criterion

From (3), the pairs of corresponding sides around angle \(Q\) are proportional:

\[\frac{QS}{QR} = \frac{QP}{QT}.\]

From (4), the included angle between these sides in both triangles is equal.

Therefore, by the SAS similarity criterion,

\[\triangle PQS \sim \triangle TQR.\]

Hence proved.