Question

E is the mid-point of a median AD of $\Delta ABC$ and BE is produced to meet AC at F. Show that $AF=\frac{1}{3}AC$.

Answer 1

Given: In $\Delta ABC$, AD is a median and E is the mid-point of AD.

Construction: Draw DP||EF.

Proof: In $\Delta ADP$, E is the mid-point of AD and EF||DP.

So, F is mid-point of AP. [by converse of mid-point theorem]

In $\Delta BFC,D$ is mid-point of BC and DP||BF.

So, P is mid-point of FC.

Thus, AF = FP = PC

$AF=\frac{1}{3}AC$ Hence proved.