In triangle $ABC$,point $D$ divides side $\overline{AC}$ so that $AD : DC = 1 : 2$. Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $BC$ and line $AE$. Given that the area of $\Delta ABC$ is 360, what is the area of $\Delta EBF$?

A. 24

B. 30

C. 32

D. 36

E. 40

1. 2019 AMC 8 官方解析

三角形$ABC$中，点$D$分$AC$使得$AD : DC = 1 : 2$．点$E$是$BD$的中点．$F$是$BC$和$AE$的交点．已知$\Delta ABC$的面积是360，那么$\Delta EBF$的面积是？

The easiest way is by Menelaus' Theorem .

$\frac{CA}{AD} \cdot \frac{DE}{EB} \cdot \frac{BF}{FC} = 1$

$\frac{BF}{CF} = \frac{1}{3}$

$\frac{AC}{CD} \cdot \frac{DB}{BE} \cdot \frac{EF}{FA} = 1$

$\frac{EF}{FA} = \frac{1}{3}$

$A_{\Delta EBF} = A_{\Delta ABC} \cdot \frac{BF}{BC} \cdot \frac{EF}{AF}$

$= 360 \times \frac{1}{4} \times \frac{1}{3} = 30$

Else to find the two ratios one needs to do more analysis by area ratios .

Answer should be the same .