2011SP-II-23

Ans: B
Note that $\Delta EFD \sim \Delta EBC$. Therefore, we have

$\begin{array}{cl}
& FD : BC \\
= & ED : EC \\
= & ED : (ED + DC) \\
= & 1 : (1 + 2) \\
= & 1 : 3
\end{array}$

Since $ABCD$ is a parallelogram, we have $AD = BC$. Hence, we have

$\begin{array}{cl}
& AF : BC \\
= & (AD – FD) : BC \\
= & (BC – FD) : BC \\
= & (3 – 1) : 3 \\
= & 2 : 3
\end{array}$