Question 626397
<pre>
{{{drawing(4000/11,400,-1,10,-3,7,

line(0,0,0,6), locate(0,0,B), locate(0.2,4.2,A), locate(0.2,6,D),
line(0,4,4.44,-1.92), locate(3,.5,C),locate(4.44,-1.92,G),
line(0,0,9,0), locate(7,0,E), locate(9,0,F),
line(7,0,4.44,-1.92)  )}}} 

The sides of &#8736;BAC are AB and AG
The sides of &#8736;GEC are EB and EG

AB&#8869;EB
AG&#8869;EG

I'll just tell you how to prove it.  You will have to write
your own two-column proof.

It's easy to prove &#5123;ABC&#8764;&#5123;EGC
because they are right triangles and 
&#8736;ACB = &#8736;FCG, so the third angles 
&#8736;BAC = &#8736;GEC.

So that's the case when they are both
acute angles.   

The reason the "or supplementary" has to be in the theorem is because:
 
1. what is true about the sides being perpendicular for acute angle
&#8736;BAC is also true for its supplementary obtuse angle &#8736;DAC.

and similarly

2. what is true about the sides being perpendicular for acute angle 
&#8736;GEC is also true for its supplementary obtuse angle &#8736;FEG.

If you have any questions, ask me in the thank-you note and I'll get
back with you.

Edwin</pre>