Question

In $\Delta ABC,AD$ is median through A and E is mid-point of AD. BE is produced to meet AC at F. Then prove that $AF=\frac{1}{3}AC.$

Answer 1

Given AD is the median of $\Delta ABC$ and E is the midpoint of AD
Through D, draw DG||BF
In $\Delta ADG$, E is the midpoint of AD AND EF||DG
By converse of midpoint theorem we have
F is midpoint of AG and AF =FG $\to \left(1\right)$
Similarly, in $\Delta BCF$
D is the midpoint of BC and DG||BF
G is midpoint of CF and FG =GC $\to \left(2\right)$
From equations (1) and (2), we get
AF =FG =GC $\to \left(3\right)$
From the figure we have, AF +FG +GC =AC
AF +AF +AF =AC [From (3)]
3 AF =AC
$AF=\left(\frac{1}{3}\right)$ AC