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

Construction Draw DG||EF.

Proof In $\Delta ADG$, E is the mid-point of AD and EF||DG.

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

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

So, G is mid-point of FC. [by converse of mid-point theorem]

Thus, $AF=FG=GC$

$\therefore AF=\frac{1}{3}AC$

Hence proved.