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...
 A: 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
$$x=\sqrt[3]2-1$$
A: Here is a proof with only Pythagoreans:
(as I see, Michael Rozenberg used also the altitude theorem of right triangles)
We have 5 eqns in 5 variables AE, AC, BA, BD, AD:
Obviously 
$AE = AC - CE = AC - 1;$
and 4 Pythagoreans:
$$
AC^2 = - AB^2 + (BD+DC)^2 = - AB^2 + (BD+1)^2;\\
AC^2 = AD^2 + DC^2 = AD^2 + 1;\\
AD^2 + BD^2 = AB^2;\\
1 = BE^2 = AB^2 + AE^2;
$$ 
Using the  third one to eliminate AB everywhere:
$$
AE = AC - 1;\\
AC^2 = - AD^2 - BD^2 + (BD+1)^2;\\
AC^2 = AD^2 + 1;\\
AD^2 + BD^2 = 1 - AE^2;
$$ 
Using the third one of this block to eliminate AD everywhere:
$$AE = AC - 1;\\
2 AC^2 = 1 - BD^2 + (BD+1)^2;\\
AC^2 + BD^2 = 2 - AE^2;
$$
Using the first  one of this block to eliminate AC everywhere:
$$2 (AE +1)^2 = 1 - BD^2 + (BD+1)^2 = 2 + 2 BD;\\
BD^2 = 2 - AE^2 - (AE + 1)^2;$$
Using the second one of this block to eliminate $BD^2$:
$$((AE +1)^2 -1 )^2 = 2 - AE^2 - (AE + 1)^2;$$
or, with $AE = x-1$,
$$ 0 = -(x^2 -1 )^2 + 2 - (x-1)^2 - x^2 = x (2 - x^3)$$
So we have the only positive real solution $AE = \sqrt[3] 2  -1$
