Question 334070
in a quadrilateral ABCD , angle B = 90°. If {{{AD^2=AB^2+BC^2+CD^2}}}, then prove that angle ACD = 90°
<pre><b>
{{{drawing(400,800/3,-.25,2.75,-.5,1.5, 

line(0,0,0,1), line(0,0,2,0), line(2,0,2.4472135955,.894427191),
line(0,1,2.4472135955,.894427191),
green(line(0,1,2,0)), rectangle(0,0,.1,.1),
locate(0,0,B), locate(0,1.15,A), locate(2,0,C), locate(2.5,1.05,D),
red(locate(1.7,.3,"90°???"), arc(2,0,1,-1,63.43494882,153.4349488)) 

 )}}}
 
We know that ABC is a right triangle because angle B = 90°
So we can use the Pythagorean theorem on that right triangle:

{{{red(AB^2 + BC^2) = AC^2}}}

We are given that 

{{{AD^2 = red(AB^2 + BC^2) + CD^2}}}

Not that the terms I've colored red are the same in both
equations so we can substitute {{{AC^2}}} for the two red
terms in the second equation,

{{{AD^2 = AC^2 + CD^2}}} 

Now we use the converse of the Pythagorean theorem which
states that the Pythagorean theorem holds ONLY for right 
triangles, thus triangle ACD is a right triangle and angle
ACD is a right angle or 90° 

Edwin</pre>