Discussion Forum

2019 AMC 8 Question 24

 
 
Picture of Ellen Zong
2019 AMC 8 Question 24
by Ellen Zong - Thursday, 9 April 2020, 1:22 AM
 

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?

 
Picture of John Lensmire
Re: 2019 AMC 8 Question 24
by John Lensmire - Thursday, 9 April 2020, 5:13 PM
 

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?