In the adjoining figure, in △ABC, AD is the median through A and E is the mid-point of AD. If BE produced meets AC in F, prove that AF = $\dfrac{1}{3}$ AC.

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.

Given,

AD is the median to BC.

BD = CD

In △BCF,

Since, D is the mid-point of BC and DG // BF, thus by converse of mid-point theorem,

DG will bisect CF, thus G is the mid-point of CF.

⇒ CG = GF ……(1)

In △ADG,

Since, E is the mid-point of AD and EF // DG, thus by converse of mid-point theorem,

EF will bisect AG, thus F is the mid-point of AG.

⇒ AF = GF ……..(2)

From (1) and (2),

⇒ AF = GF = CG

From figure,

⇒ AC = AF + GF + CG

⇒ AC = AF + AF + AF

⇒ AC = 3 AF

⇒ AF = $\dfrac{1}{3}$ AC.

Hence, proved that AF = $\dfrac{1}{3}$ AC.