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

Question 3

Given: In Fig. 6.18, LM \(\parallel\) CB and LN \(\parallel\) CD.

1. Use LM \(\parallel\) CB

Consider \(\triangle ABC\) with points \(M\) on \(AB\) and \(L\) on \(AC\), and \(LM \parallel CB\).

Then,

\(\angle AML = \angle ABC\) (corresponding angles)

\(\angle ALM = \angle ACB\) (corresponding angles)

So, \(\triangle AML \sim \triangle ABC\) (AA similarity).

From similarity, corresponding sides are proportional:

\[\frac{AM}{AB} = \frac{AL}{AC} \quad ...(1)\]

2. Use LN \(\parallel\) CD

Now consider \(\triangle ADC\) with points \(N\) on \(AD\) and \(L\) on \(AC\), and \(LN \parallel CD\).

Then,

\(\angle ANL = \angle ADC\) (corresponding angles)

\(\angle ALN = \angle ACD\) (corresponding angles)

So, \(\triangle ALN \sim \triangle ADC\) (AA similarity).

Again, corresponding sides are proportional:

\[\frac{AN}{AD} = \frac{AL}{AC} \quad ...(2)\]

3. Compare the two ratios

From (1): \(\dfrac{AM}{AB} = \dfrac{AL}{AC}\)

From (2): \(\dfrac{AN}{AD} = \dfrac{AL}{AC}\)

Since both are equal to \(\dfrac{AL}{AC}\), we get

\[\frac{AM}{AB} = \frac{AN}{AD}.\]

Hence proved.