Question

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

• 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 get

$AF=FG=GC$........ (3)

From the figure we have,$AF+FG+GC=AC$

$AF+AF+AF=AC$ [from (3)]

$3AF=AC$

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

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