Question

In the given figure, $AD$ is a median of $△ABC$ and $E$ is the mid-point of $AD.$ Also, $BE$ produced meets $AC$ in $F.$ Then $AF=$

• A. $3AC$
• B. $\frac{1}{2}AC$
• C. $2AC$
• D. $\frac{1}{3}AC$

Answer 1

The correct option is D

$\frac{1}{3}AC$

Draw $DP\parallel EF$

In $\Delta ADP$, $E$ is the mid-point of $AD$ and $EF\parallel DP$
$\therefore$ $F$ is the mid-point of $AP.$
In $\Delta FBC$, $D$ is the mid-point of $BC$ and $DP\parallel BF$.
$\therefore$ $P$ is the mid-point of $FC$
Thus, $AF=FP=PC$
Hence, $AF=\frac{1}{3}AC$