2011SP-II-23

Ans: B
Note that

Δ E F D \sim Δ E B C

. Therefore, we have

$\begin{array}{cl} F D : B C \\ = & E D : E C \\ = & E D : (E D + D C) \\ = & 1 : (1 + 2) \\ = & 1 : 3 \end{array}$

Since $A B C D$ is a parallelogram, we have $A D = B C$ . Hence, we have

$\begin{array}{cl} A F : B C \\ = & (A D - F D) : B C \\ = & (B C - F D) : B C \\ = & (3 - 1) : 3 \\ = & 2 : 3 \end{array}$

2011SP, HKDSE-MATH, Paper 2 Mensuration