Help with simple internal line segments of triangles proof.

**jpatrie** · February 7th 2011, 11:06 AM

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

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.

**Ithaka** · February 7th 2011, 11:23 AM

Originally Posted by jpatrie

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

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:

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

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

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

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

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

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