NCERT Solutions
Class 10 - Mathematics - Chapter 6: TRIANGLES - Exercise 6.2

Question 6

Given: In \(\triangle PQR\), point \(O\) lies inside the triangle. Points \(A\), \(B\) and \(C\) lie on segments \(OP\), \(OQ\) and \(OR\) respectively, such that \(AB \parallel PQ\) and \(AC \parallel PR\).

To prove: \(BC \parallel QR\).

Step 1: Use AB \(\parallel\) PQ in \(\triangle POQ\)

In \(\triangle POQ\), with \(A\) on \(OP\), \(B\) on \(OQ\) and \(AB \parallel PQ\), by the Basic Proportionality Theorem,

\[\frac{OA}{OP} = \frac{OB}{OQ} \quad ...(1)\]

Step 2: Use AC \(\parallel\) PR in \(\triangle POR\)

In \(\triangle POR\), with \(A\) on \(OP\), \(C\) on \(OR\) and \(AC \parallel PR\), again by the Basic Proportionality Theorem,

\[\frac{OA}{OP} = \frac{OC}{OR} \quad ...(2)\]

Step 3: Relate the ratios on OQ and OR

From (1) and (2), the left-hand sides are equal, so

\[\frac{OB}{OQ} = \frac{OC}{OR} \quad ...(3)\]

Step 4: Apply the converse of Basic Proportionality Theorem in \(\triangle QOR\)

In \(\triangle QOR\), points \(B\) and \(C\) lie on \(OQ\) and \(OR\) respectively, and from (3) we have

\[\frac{OB}{OQ} = \frac{OC}{OR}.\]

By the converse of the Basic Proportionality Theorem, if a line cuts two sides of a triangle in the same ratio, then that line is parallel to the third side.

Therefore, the line joining \(B\) and \(C\) is parallel to \(QR\):

\[BC \parallel QR.\]

Hence proved.