<s>[INST] 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 
 [/INST] 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, $\gamma=2\alpha \Rightarrow \measuredangle CDE=2\measuredangle ABE$
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