Question 1034944
{{{drawing(320,400,-.5,3.5,-.5,4.5,

triangle(0,0,3,0,0,4),
triangle(0,0,0,2.5,1,0), 
triangle(0,4,0,0,1,0),
triangle(0,0,0,2.5,3,0),rectangle(0,0,.1,.1),
locate(0,0,B), locate(3,0,C),locate(-.14,2.6,D),
locate(1,0,E),locate(0,4.2,A)   )}}}
<pre>
&#916;ABC is a right triangle, so by the Pythagorean
theorem, 

(1)     AB˛+BC˛ = AC˛


&#916;ABE is a right triangle, so by the Pythagorean
theorem, 

(2)     AB˛+BE˛ = AE˛.

&#916;DBC is a right triangle, so by the Pythagorean
theorem, 

(3)     DB˛+BC˛ = CD˛.

&#916;DBE is a right triangle, so by the Pythagorean
theorem, 

(4)     DB˛+BE˛ = DE˛.

Add BE˛+DB˛  to both sides of equation (1)


        AB˛+BC˛+BE˛+DB˛ = AC˛+BE˛+DB˛

Rearrange the terms:

        AB˛+BE˛+DB˛+BC˛ = AC˛+DB˛+BE˛

To make things easier to see, let's put parentheses 
around the first two terms on the left, the last two 
terms on the left, and the last two terms on the right:

(5)   (AB˛+BE˛)+(DB˛+BC˛) = AC˛+(DB˛+BE˛)
 
Using (2) we replace (AB˛+BE˛) in (5) by AE˛ 

Using (3) we replace (DB˛+BC˛) in (5) by CD˛

Using (4) we replace (DB˛+BE˛) in (5) by DE˛

And we end up with what we were to prove:

       AE˛+CD˛ = AC˛+DE˛
       
Edwin</pre>