<s>[INST] In right triangle $ABC$ ($\angle A=90$), $E$ is a point on $AC$.Find $AE$ given that... 
In right triangle $ABC$ ($\angle A=90$), $AD$ is a height and $E$ is a point on $AC$ so that $BE=EC$ and $CD=CE=EB=1$. $\color {red} {Without}$ using trigonometric relations find  $AE$.


I do have a solution USING trigonometric relations ( $AE=\sqrt[3]{2}-1$ ),but it seems other solutions are troublesome.My attempt led to a complicated polynomial equation...
 [/INST] Let $AE=x$.
Hence, $AD=\sqrt{(x+1)^2-1}=\sqrt{x^2+2x}$ and since $AD^2=BD\cdot DC$,
we obtain $BD=x^2+2x$, which says that $AB=\sqrt{(x^2+2x)^2+x^2+2x}$ and
$$BE=\sqrt{(x^2+2x)^2+x^2+2x+x^2}.$$ 
Thus,$$(x^2+2x)^2+x^2+2x+x^2=1$$ or
$$x^4+4x^3+6x^2+2x-1=0$$ or
$$(x+1)(x^3+3x^2+3x-1)=0$$ or
$$(x+1)^3=2$$ or
$$x=\sqrt[3]2-1$$
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