Zoom International Math League: 2019 AMC 8 Question 24

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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?

First to clarify the problem, it should be $A D : D C = 1 : 2$ , 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 $[A B C]$ denote the area of $△ A B C$ so $[A B C] = 360$ . Since $A D : A C = 1 : 2$ triangles $△ A B D$ (with base $A D$ ) and $△ C B D$ (with base $C D$ ) share a height, so their ratio of areas $[A B D] : [C B D] = 1 : 2.$ Since the areas add up to $360$ , we have $[A B D] = 120$ and $[C B D] = 240$ .

Using this type of reasoning is key to the problem. One nice way to proceed is to let $G$ be on $B C$ so that $\overset{―}{E G}$ is parallel to $\overset{―}{A C}$ . What similar triangles can you find from here?

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