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

Question 4

Given: In \(\triangle ABC\), points \(D\) lies on \(AB\), and \(E, F\) lie on \(BC\). Lines are drawn so that \(DE \parallel AC\) and \(DF \parallel AE\).

To prove: \(\dfrac{BF}{FE} = \dfrac{BE}{EC}\).

Step 1: Use similarity in \(\triangle ABC\)

Since \(DE \parallel AC\) and \(D\) is on \(AB\), \(E\) is on \(BC\), we have:

\[\triangle BDE \sim \triangle BAC \quad (\text{AA similarity})\]

So the corresponding sides are proportional:

\[\frac{BD}{BA} = \frac{BE}{BC} = \frac{DE}{AC} \quad \text{...(1)}\]

Step 2: Use similarity in \(\triangle ABE\)

Since \(DF \parallel AE\) and \(D\) is on \(AB\), \(F\) is on \(BE\), we have:

\[\triangle BDF \sim \triangle BAE \quad (\text{AA similarity})\]

Thus,

\[\frac{BD}{BA} = \frac{BF}{BE} = \frac{DF}{AE} \quad \text{...(2)}\]

Step 3: Relate \(BF\) and \(BE\) using (1) and (2)

From (1) and (2), the common ratio is \(\dfrac{BD}{BA}\). Hence,

\[\frac{BE}{BC} = \frac{BD}{BA} = \frac{BF}{BE}\]

So,

\[\frac{BF}{BE} = \frac{BE}{BC} \quad \text{...(3)}\]

Step 4: Express segments on \(BC\)

Points \(B, F, E, C\) lie on the same straight line in that order, so:

\[BC = BE + EC, \quad BE = BF + FE\]

From (3):

\[BF \cdot BC = BE^2\]

Substitute \(BC = BE + EC\):

\[BF(BE + EC) = BE^2\]

\[BF \cdot BE + BF \cdot EC = BE^2\]

But \(BE = BF + FE\), so:

\[BE^2 = BE(BF + FE) = BF \cdot BE + BE \cdot FE\]

Comparing with the previous expression:

\[BF \cdot BE + BF \cdot EC = BF \cdot BE + BE \cdot FE\]

Cancel \(BF \cdot BE\) from both sides:

\[BF \cdot EC = BE \cdot FE\]

Step 5: Form the required ratio

Divide both sides by \(FE \cdot EC\) (non-zero):

\[\frac{BF}{FE} = \frac{BE}{EC}\]

Hence proved.