Question

In the adjoining figure a $\Delta$ $ABC$ has been given in which $AD$ is its median $E$ is the midpoint of $AD,BE$ produced meets $AC$ in $F$ and $DGIEF$ meets $AC$ at $G$. If $AC=5.4$ cm then the length of $AF$ is:

• A. $3.6$ cm
• B. $2.7$ cm
• C. $1.8$ cm
• D. $10.8$ cm

Answer 1

Answer: (C)

$1.8$ cm

in $\Delta$ $ADG$, $E$ is the midpoint of $AD$ and $EF||DG$
So $EF$ must bisect $AG$.
$\therefore$ $F$ is the midpoint of $AG$
So, $AF=FG$
In $\Delta$ $BCF$, $D$ is the midpoint $BC$ and $DG||BF$

So, $DG$ must bisect $FC$
$\therefore$ $G$ is the midpoint $FC$
So, $FG=GC$
Thus $AF=FC=GC$
$AF=\frac{1}{3}\times AC=\frac{1}{3}\times 5.4=1.8$ cm