Question

In the given figure, AD is a median of ABC and E is the mid-points 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 $=\frac{1}{3}AC$