NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 6: Triangles - Exercise 6.3
Question 3

Question. 3

In Fig. 6.9, if \(\angle 1=\angle 2\) and \(\triangle NSQ \cong \triangle MTR\), prove that \(\triangle PTS \sim \triangle PRQ\).

Fig. 6.9: Large triangle PNR with base MQRN, cevians through S and T to Q and R.

Answer:

Proved.

Detailed Answer with Explanation:

Step 1: It is given that \(\triangle NSQ \cong \triangle MTR\).

From congruence, we know that corresponding angles and sides are equal. So, \(\angle SQN = \angle TRM\) and \(SQ = TR\).

Step 2: Points \(M, Q, R, N\) are collinear (they lie on a straight line). Using this and the equal angles \(\angle SQN = \angle TRM\), we can say that line segment \(QS \parallel RT\).

Step 3: It is also given that \(\angle 1 = \angle 2\). These are the angles at point \(P\), so we can write: \(\angle SPT = \angle QPR\).

Step 4: Because \(QS \parallel RT\):

  • \(\angle PST = \angle PRQ\) (alternate interior angles)
  • \(\angle PTS = \angle PQR\) (corresponding angles)

Step 5: Now compare triangles \(PTS\) and \(PRQ\):

  • \(\angle SPT = \angle QPR\)
  • \(\angle PST = \angle PRQ\)

So, two pairs of corresponding angles are equal.

Step 6: By the AA similarity criterion, if two angles of one triangle are equal to two angles of another triangle, then the triangles are similar.

Therefore, \(\triangle PTS \sim \triangle PRQ\).

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