Angle of a triangle inscribed in a square Say we have a square $ABCD$. Put points $E$ and $F$ on sides $AB$ and $BC$ respectively, so that  $BE = BF$. Let $BN$ be the altitude in triangle $BCE$. What is $\angle DNF$?
I'm inclined to say that it's a right angle because that's what it looks like from what I've drawn, but I have no idea how to proceed. 



Firstly We can write
(1) Oklid relation from $\triangle CNB$ 
$h^2=m(k+x)$  
(2) Pisagor relation from $\triangle DPN$ 
$a^2=(x+k)^2+(x+k+m-h)^2$   
(3) Pisagor relation from $\triangle FRN$
$b^2=h^2+k^2$
(4) Pisagor relation from $\triangle DCF$ 
$c^2=x^2+(x+k+m)^2$

if DNF triange is a right triangle,It must satisfy $a^2+b^2=c^2$. 
$(x+k)^2+(x+k+m-h)^2+h^2+k^2=x^2+(x+k+m)^2$ 
$(x+k)^2+(x+k+m)^2-2h(x+k+m)+h^2+h^2+k^2=x^2+(x+k+m)^2$
$(x+k)^2-2h(x+k+m)+2h^2+k^2=x^2$
$xk+k^2-h(x+k+m)+h^2=0$
$xk+k^2-h(x+k+m)+m(k+x)=0$
$-h(x+k+m)+(k+m)(k+x)=0$
$\frac{x+k}{x+k+m}=\frac{h}{k+m}$
This result is equal to the rates of thales formula for similar triangles   $\triangle CRN \sim \triangle CBE$  
Thus $a^2+b^2=c^2$  is correct for $\triangle DNF$ .
