Finding segment in a right triangle. Here is the picture of the question:


  
*
  
*$ABC$ is a right triangle.
  
*$m(CBA)=90^\circ$.
  
*$m(BAD)=2m(DAC)=2\alpha$.
  
*$D$ is a midpoint of $[BC]$.
  
*$E$ is a point on $[AD]$.
  
*$m(BED)=90^\circ$.
  
*$|DE|=3$.
  
*What is $|AB|=x$?
  

Tried lots of things which gives me some trigonometric identities, but none of them led me to the solution which is $x=6$. And even if it did, i prefer more geometric methods (all solutions are welcome though). Note that this could be an easy question, and i could probably missing something obvious.
 A: This is one approach.  I don't pretend it's the nicest.
Let $u = \tan\alpha$.  Then
$$
\tan2\alpha = \frac{2u}{1-u^2}
$$
$$
\tan3\alpha = \frac{u(3-u^2)}{1-3u^2}
$$
Since $\tan3\alpha = 2\tan2\alpha$, we have
$$
\frac{4u}{1-u^2} = \frac{u(3-u^2)}{1-3u^2}
$$
$$
(1-u^2)(3-u^2) = 4(1-3u^2)
$$
$$
3-4u^2+u^4 = 4-12u^2
$$
$$
1-u^4 = 8u^2
$$
Now let $y = BE$.  By similar triangles, we have
$$
\frac{3}{y} = \frac{\sqrt{9+y^2}}{x} = \tan2\alpha = \frac{2u}{1-u^2}
$$
Square all terms to obtain
$$
\frac{9}{y^2} = \frac{9+y^2}{x^2} = \left(\frac{2u}{1-u^2}\right)^2
                                  = \frac{4u^2}{1-2u^2+u^4}
$$
Observe that the outer pair of the equality above gives us
$$
\frac{9+y^2}{y^2} = \frac{9}{y^2}+1 = 1+\frac{4u^2}{1-2u^2+u^4}
                                    = \frac{1+2u^2+u^4}{1-2u^2+u^4}
                                    = \left(\frac{1+u^2}{1-u^2}\right)^2
$$
Multiplying both ends by $\frac{x^2}{9+y^2} \cdot \frac{y^2}{9} = \left(\frac{1-u^2}{2u}\right)^2\left(\frac{1-u^2}{2u}\right)^2$ yields 
\begin{align}
\frac{x^2}{9} & = \left(\frac{1+u^2}{2u}\right)^2
                  \left(\frac{1-u^2}{2u}\right)^2 \\
              & = \left(\frac{1-u^4}{4u^2}\right)^2 \\
              & = \left(\frac{8u^2}{4u^2}\right)^2
                  \qquad \longleftarrow 1-u^4 = 8u^2 \\
              & = 2^2 = 4
\end{align}
So $x^2 = 36$ and $x = 6$.
Neat question.  One almost feels as though the simple and straightforward answer must be obtainable with a correspondingly simple and straightforward approach.  Thus far, though, I haven't seen it.  Anybody?
A: 
(1) The brown dotted line is the extension of $BE$.
(2) The green dotted line ($AC’$) is the angle bisector of $∠BAE$ such that $\beta_1 = \beta_2 = \beta$. Another obvious fact is $\beta ‘ = 2\beta$.
(3) The blue dotted line is the perpendicular bisector of $AB$ cutting $AB$ and $AC$ at $P$ and $Q$ respectively.
By intercept theorem, $AQ = QC$.
By midpoint theorem, $QD = BP = PA$.
$L$ is point on $BC$ such that $EL \bot BC$. $EL$ is extended to cut $AC’$ at $N$. It should also be clear that $\angle 1 = \angle 2 = \beta’$..
The line $AEDM$ is the axis of symmetry for $\triangle ACC’$. Then, $\angle AMC = 90^\circ$. This means M is also on the circum-circle (centered at Q) of $\triangle ABC$.
Therefore, $\triangle BED \equiv  \triangle CMD \Rightarrow DE = DM$. 
$QA = QM \Rightarrow \beta_3 = \beta$. $\beta_4 = \angle 2 - \beta_3 = \beta \Rightarrow QD = DM$.
Result follows.
