How prove $\measuredangle CDE=2\measuredangle ABE$ In rectangular  $ABCD$,and $E\in AC$,such
$$BE=\sqrt{2}\cdot AE$$
show that
$$\measuredangle CDE=2\measuredangle ABE$$

My try: let $$AB=a,AD=b,\dfrac{AE}{AC}=k,$$
then
$$AE=k\sqrt{a^2+b^2},BE=k\sqrt{2(a^2+b^2)}$$
I know have this nice relsut
$$AE^2+EC^2=BE^2+ED^2$$
then
$$ED^2=k^2\cdot AC^2+(1-k)^2\cdot AC^2-2k^2\cdot AC^2=(1-2k)AC^2$$
so
$$\cos{\measuredangle EBD}=\dfrac{AB^2+BE^2-AE^2}{2AB\cdot BE}=\dfrac{a^2+k^2(a^2+b^2)}{2a\cdot k\sqrt{2(a^2+b^2)}}$$
$$\cos{\measuredangle CDE}=\dfrac{DC^2+DE^2-EC^2}{2 DC\cdot DE}=\dfrac{a^2+(1-2k)(a^2+b^2)-(1-k)^2(a^2+b^2)}{2\sqrt{(1-2k)(a^2+b^2)}a}=\dfrac{a^2-k^2(a^2+b^2)}{2\sqrt{(1-2k)(a^2+b^2)}a}$$
we will prove 
$$2\cos^2{\measuredangle ABE}-1=\cos{\measuredangle CDE}$$
$$\Longleftrightarrow 2\dfrac{((k^2+1)a^2+b^2)^2}{8a^2k^2(a^2+b^2)}-1=\dfrac{a^2-k^2(a^2+b^2)}{2\sqrt{(1-2k)(a^2+b^2)}a}$$
But I can't
Thank you for you 
 A: Probably not the best way to do it, but maybe you can simplify it.
Let $F \in AB$ such that $AB\perp EF$ and let $G \in CD$ such that $CD \perp EG$.
Let $|AE|=x\Rightarrow |BE|=x\sqrt2$
Let $|EG|=y$
Finally let $\measuredangle ABE=\alpha$, $\measuredangle BAC = \measuredangle ACD = \beta$ and $\measuredangle ADC=\gamma$
$\sin\alpha=\Large\frac{|FE|}{x\sqrt2}$ $\Rightarrow |FE|=\sin\alpha\cdot x\sqrt2$
Likewise, $|BF|=|CG|=\cos\alpha\cdot x\sqrt2$ and $|AF|=|GD|=\cos\beta\cdot x$
$\sin\beta=\Large\frac{|EF|}{|AE|}$$=\sin\alpha\cdot \sqrt2$
$\tan\beta = \Large\frac{y}{\cos\alpha\cdot x\sqrt2}$=$\Large\frac{\sin\beta}{\cos\beta}$ $\Rightarrow y=\Large\frac{\sin\beta\cdot \cos\alpha\cdot x\sqrt2}{\cos\beta}$
Since $\cos\beta=\sqrt{1-\sin^2\beta}=\sqrt{1-2\sin^2\alpha}$ and $\sin\beta=\sin\alpha\cdot \sqrt2$ we can plug these in. So,
$y=\Large\frac{\sin\alpha\cdot \cos\alpha\cdot 2x}{\sqrt{1-2\sin^2\alpha}}$
$\tan\gamma=\Large\frac{y}{\cos\beta\cdot x}$=$\Large\frac{y}{\sqrt{1-2\sin^2\alpha}\cdot x}$
Plug $y$ in
$\tan\gamma=\Large\frac{\sin\alpha\cdot \cos\alpha\cdot 2x}{(1-2\sin^2\alpha)\cdot x}$=$\Large\frac{2\sin\alpha\cos\alpha}{1-2\sin^2\alpha}$=$\Large\frac{\sin2\alpha}{\cos2\alpha}$=$\tan2\alpha$
Hence, $\gamma=2\alpha \Rightarrow \measuredangle CDE=2\measuredangle ABE$
A: I solve this problem in another way. Let us establish a coordinate system which $A$ is original point and $AB$ is y-axis and $AD$ is x-axis. And $AB=a$, $AD=b$.
In this frame, we have: $A(0,0)$, $B(0,a)$, $C(b,a)$, $D(b,0)$. For the sake of simplicity, we define $k:=\frac{b}{a}$ and equation of $AC$ can be written as $y=kx$. Also, assume $E(bx_0,ax_0)$ where $x_0$ is a parameter to be determined. And then the length of $AE$ and $BE$ can be written as :
\begin{equation}
AE=\sqrt{1+k^2}bx_0\\
BE=b\sqrt{x_0^2+k^2(1-x_0)^2}
\end{equation}
Note that the condition $BE=\sqrt{2}AE$ gives:
\begin{equation}
\frac{BE}{AE}=\sqrt{\frac{x_0^2+k^2(1-x_0^2)}{x_0^2+k^2x_0^2}}=\sqrt{2}\\
(1+k^2)x_0^2+2k^2x_0-k^2=0\\
x_0=\frac{k(-k+\sqrt{2k^2+1})}{k^2+1}\quad\mbox{(The negative root have been ignored since $E\in AC$)}
\end{equation}
Remark: I should point out the this result have been ignored in your deduction that the point $E$ can be determined by $a$ and $b$. 
Let $\alpha$ and $\beta$ denote $\measuredangle ABE$ and $\measuredangle EDC$ respectively. We have:
\begin{equation}
\cos\alpha=\frac{AB^2+BE^2-AE^2}{2AB\cdot BE}=\frac{a^2+AE^2}{2\sqrt{2}a\cdot AE}\\
\cos\beta=\frac{ED^2+CD^2-EC^2}{2ED\cdot CD}=\frac{ED^2+a^2-EC^2}{2a\cdot ED}
\end{equation}
Now, the coordinate of $E$ is governed by $a$ and $b$. $AE$, $ED$ and $EC$ can be computed by coordinates of $A$, $B$, $C$, $D$, $E$ and all of them are functions of $k$. By a direct computation, we can check:
\begin{equation}
\cos\beta=2\cos^2\alpha-1
\end{equation}
Since the deduction is so long that I ignore them here. You can check it by using MAPLE if you feel it is hard.
