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

**Solution:**

We know according to the basic proportionality theorem, if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.

In ΔABC

DE || AC

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

In ΔABE

DF || AE

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

From (i) and (ii)

BD/AD = BE/EC = BF/FE

Thus, BE/EC = BF/FE

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 6

**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

**Summary:**

In the figure, if DE || AC and DF || AE we have proved that BF/FE = BE/EC.

**☛ Related Questions:**

- In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
- In Fig. 6.21, A, B and C are points on OP, OQ, and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
- Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
- Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

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