Question

In $△ABC,AD$ is median and E is the mid-point of AD. If BE is extended. it meets AC in F.
AB=8 cm. BC = 21 cm and AC = 15 cm. then AF=

• A. 7 cm
• B. 3 cm
• C. 12 cm
• D. 5 cm

Answer 1

The correct option is D 5 cm

Given

$AD$ is the median of $△ABC$ and $E$ is the midpoint of $AD$

Through $D$

draw $DG\parallel BF$

In $△ADG$

$E$ is the midpoint of$AD$ and $EF\parallel DG$

By converse of midpoint theorem we have

$F$ is midpoint of $AG$ and $AF=FG$ ..............1

Similarly, in $△BCF$

$D$ is the midpoint of $BC$ and $DG\parallel BF$

$G$ is midpoint of $CF$ and $FG=GC$ ..............2

From equations 1 and 2

we will get

$AF=FG=GC$ ........3

$AF+FG+GC=AC$

$AF+AF+AF=AC$ (from eu 3)

$3AF=AC$

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

$\therefore AF=\frac{1}{3}\times 15=5cm$