# Thread: Help with simple internal line segments of triangles proof.

1. ## Help with simple internal line segments of triangles proof.

Prove that if $\displaystyle D$ is internal to $\displaystyle \overline{AB}$, $\displaystyle E$ is internal to $\displaystyle \overline{AC}$ and $\displaystyle \frac{DB}{ DA} = \frac{EC}{EA}$ then $\displaystyle {BC}$ is parallel to $\displaystyle {DE}$

I'm having a tough time figuring out where to start or how to go about doing this proof, we've been focusing on Ceva's theorem as of late so I'm thinking I have to incorporate it some how.

2. Originally Posted by jpatrie
Prove that if $\displaystyle D$ is internal to $\displaystyle \overline{AB}$, $\displaystyle E$ is internal to $\displaystyle \overline{AC}$ and $\displaystyle \frac{DB}{ DA} = \frac{EC}{EA}$ then $\displaystyle {BC}$ is parallel to $\displaystyle {DE}$

I'm having a tough time figuring out where to start or how to go about doing this proof, we've been focusing on Ceva's theorem as of late so I'm thinking I have to incorporate it some how.
First re-write the given equality:

$\displaystyle \frac{DB}{ DA} = \frac{EC}{EA}$

$\displaystyle \frac{DB+DA}{ DA} = \frac{EC+EA}{EA}$

$\displaystyle \frac{AB}{AD} = \frac{AC}{AE}$

$\displaystyle \frac{AD}{AB} = \frac{AE}{AC}$

Then look at the triangles ABC and ADE. What can you say about them?