AD and BE are two medians of a ABC. F is a point on AC such that DF || BE. If AC = 12 cm, then FC =

1. 6 cm

2. 4 cm

3. 3 cm

4. none of these

In △BCE,

D is the midpoint of BC (As AD is median)

Given,

DF || BE

By converse of mid-point theorem,

A line drawn through the midpoint of one side of a triangle, and parallel to another side, will bisect the third side.

Thus, in triangle BEC,

F is the mid-point of CE

⇒ FC = $\dfrac{1}{2}$ CE ….(1)

Given,

BE is median.

⇒ E is the mid-point of AC

⇒ AE = CE

⇒ CE = $\dfrac{1}{2}$ AC

Substituting value of CE in eq.(1), we get:

⇒ FC = $\dfrac{1}{2} \times \dfrac{1}{2} \text{AC}$

⇒ FC = $\dfrac{1}{4} \times 12$

⇒ FC = 3 cm.

Hence, option 3 is the correct option.