In the given figure, AD and CE are medians and DF // CE. Prove that : FB = $\dfrac{1}{4}AB$.

By mid-point theorem,

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

By converse of mid-point theorem,

The straight line drawn through the mid-point of one side of a triangle parallel to another, bisects the third side.

Since, AD and CE are medians.

∴ D is the mid-point of BC and E is the mid-point of AB.

In △ BEC,

DF || CE and D is the mid-point of BC.

∴ F is the mid-point of BE. (By converse of mid-point theorem)

∴ FB = $\dfrac{1}{2}BE$ …….(1)

Since, E is the mid-point of AB.

∴ BE = $\dfrac{1}{2}AB$ …….(2)

Substituting value of BE from equation (2) in (1), we get :

∴ FB = $\dfrac{1}{2} \times \dfrac{1}{2} \times AB = \dfrac{1}{4}AB$.

Hence, proved that FB = $\dfrac{1}{4}AB$.