I'm trying to answer the problem below using similarity of triangles, but wherever I go I got a fourth degree equation that is quite hard to solve. There's another way to solve this? I need a hint. Thanks!
I saw that $FD=\sqrt{1-x^2}$ then I did
$BA/AE=FB/DE\Rightarrow 1/(1+x)^2=\sqrt{1-x^2}/x\Rightarrow0-x^4-2x^3-x^2+2x+1$
$$\frac{x}{1+x}=\frac{1}{\sqrt{1+(1+x)^2}}$$ or $$x^4+2x^3+x^2-2x-1=0$$ or $$x^4+2x^3+3x^2+2x+1-2x^2-4x-2=0$$ or $$(x^2+x+1)^2-2(x+1)^2=0$$ or $$(x^2+x+1+\sqrt2(x+1))(x^2+x+1-\sqrt2(x+1))=0$$ or $$(x^2+(\sqrt2+1)x+1+\sqrt2)(x^2-(\sqrt2-1)x+1-\sqrt2)=0,$$ which gives $$x=\frac{\sqrt2-1+\sqrt{2\sqrt2-1}}{2}.$$