## Discussion Forum

### 2019 AMC 8 Question 24

2019 AMC 8 Question 24

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

Re: 2019 AMC 8 Question 24

First to clarify the problem, it should be $AD : DC = 1:2$, right?

There are many different solutions here, but most rely on using the ratio of areas for triangles when they share a base or share a height.

For example, let $[ABC]$ denote the area of $\triangle ABC$ so $[ABC] = 360$. Since $AD:AC = 1:2$ triangles $\triangle ABD$ (with base $AD$) and $\triangle CBD$ (with base $CD$) share a height, so their ratio of areas$$[ABD] : [CBD] = 1 : 2.$$Since the areas add up to $360$, we have $[ABD] = 120$ and $[CBD] = 240$.

Using this type of reasoning is key to the problem. One nice way to proceed is to let $G$ be on $BC$ so that $\overline{EG}$ is parallel to $\overline{AC}$. What similar triangles can you find from here?