# In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC

**Solution:**

As we know if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

In DABC

DE || AC

BD/AD = BE/EC .........(i)

In DABE

DF || AE

BD/AD = BF/FE ........(ii)

From (i) and (ii)

BD/AD = BE/EC = BF/FE

BE/EC = BF/FE

**Video Solution:**

## In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC

### Class 10 Maths NCERT Solutions - Chapter 6 Exercise 6.2 Question 4:

In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC

Hence proved BF/FE = BE/EC if in the above figure DE || AC and DF || AE