Proportion Proof Problem

[Despicable]

Given: Parallelogram ABCD
Prove: TE x ER = SE x EV

The diagram is here..
http://offtype.net/image_5138582582052.gif.html
I know that I have to find TE/SE = EV/ER. But I can't seem to prove triangle TSE similar to triangle REV. And so I don't know how to prove those sides in proportion. Am I suppose to use something with a trapezoid? I'm not sure..

Please help. Thank you in advance!

 

[mmm4444bot]

… I have to [show that] TE/SE = EV/ER. But I can't seem to prove triangle TSE similar to triangle REV. And so I don't know how to prove those sides in proportion …

If you show that triangle TSE is similar to triangle REV, then you've already shown that their sides are in proportion.

In other words, once we know that two triangles are similar, we do not need to do any additional work to show that their corresponding sides are proportional because this proportionality is part of the definition of similarity.

Have you learned the basic geometry theorems that deal with things like vertical angles and alternate angles that occur when two lines intersect or when a line transverses two parallel lines?

If you're not sure, then CLICK HERE for more information.

Angles TAE and VCE are equal. Do you see why?

Angles TEA and VEC are equal. Do you see why?

This tells us that triangles ATE and CVE are similar.

The same reasoning shows that triangles ERC and EST are similar.

This approach is one of many.