答案:B
留意 $\Delta EFD \sim \Delta EBC$。所以,可得
留意 $\Delta EFD \sim \Delta EBC$。所以,可得
$\begin{array}{cl}
& FD : BC \\
= & ED : EC \\
= & ED : (ED + DC) \\
= & 1 : (1 + 2) \\
= & 1 : 3
\end{array}$
由於 $ABCD$ 為一平行四邊形,所以 $AD = BC$。由此,可得
$\begin{array}{cl}
& AF : BC \\
= & (AD – FD) : BC \\
= & (BC – FD) : BC \\
= & (3 – 1) : 3 \\
= & 2 : 3
\end{array}$