TITLE: An Interesting Question about Pythagorean Triples
QUESTION [6 upvotes]: I have recently thought about a interesting question about Pythagorean Triples.


Consider such a right-angled trapezium formed by 3 right-angled triangle. Determine does it exist integral solutions for lengths of sides $AB, BC, CD, DE, EA, AC$ and $CE$.

Here's my ideas.

I know that Pythagorean Triple can be generated by substituting integer into $x^2-y^2, 2xy, x^2+y^2.$ 
So let $AB=m^2-n^2$, $BC=2mn$, $CD=2pq$, $DE=p^2-q^2$, $AC=m^2+n^2=u^2-v^2$, $CE=p^2+q^2=2uv$ and $ AE=u^2+v^2.$ To answer the question, I have to show that if $m^2+n^2=u^2-v^2$ and $p^2+q^2=2uv$ have integral solution(*). But I don't know how to show this.
Can anyone tell me if I'm right about (*)? If I'm right, how to show it?
If I'm wrong, how to solve the question?
Thank you.
Sorry, I'm a poor question-tagger.

REPLY [9 votes]: Note that triangles ABC and CDE must be similar if BCD is to be a straight line, so both must be built from integer multiples of the same primitive pythagorean triple T= {a,b,c} with hypotenuse c. 
Suppose ABC is $pT$ and CDE is $qT$ then $AC=pc$ and $CE=qc$ and $AE^2=p^2c^2+q^2c^2$ so that $AE$ is divisible by $c$ and equals $rc$.
You can build your trapezium from any pair of pythagorean triples (not necessarily primitive) 
T={$a,b,c$} and U= {$p,q,r$} 
with ABC = {$pa,pb,pc$}, CDE = {$qb,qa,qc$}, ACE = {$pc,qc,rc$}.
And this is essentially the only way of doing it.