
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 $\triangle DNF$ .