Why am I getting two different answer for this geometry question? 
There is a right triangle $\textrm{ABC}$ like the diagram above, and the point $\textrm{D}$ set so that $\mathrm{\overline{AD}=\overline{BC}}$. If point $\mathrm{E}$ divides line segment $\mathrm{AB}$ in the ratio of $5:2$, $\mathrm{\overline{AD}=\overline{DE}=\overline{CE}}$. Let $\angle\mathrm{ABC}=\theta$, then find the value of $\tan\theta$.
So what I did was I selected a point $\mathrm{F}$ on $\mathrm{\overline{BC}}$ so that $\mathrm{\overline{EF}\perp\overline{BC}}$, then I let $\mathrm{\overline{EF}}=h$, $\mathrm{\overline{AD}}=x$.
Then I can say
$$h:h+x=2:5$$
So I get
$$h=\frac{2}{3}x$$
Then I can use pythagorean theorem on $\triangle\mathrm{EFC}$
I get $$\mathrm{\overline{FC}}=\frac{\sqrt{5}}{3}x$$
Therefore $$\tan\theta=\frac{\mathrm{\overline{FE}}}{\mathrm{\overline{BF}}}=\frac{2}{3-\sqrt{5}}=\frac{3+\sqrt{5}}{2}$$
However, If I just find $$\tan\theta=\frac{\mathrm{\overline{CA}}}{\mathrm{\overline{BC}}}=\frac{7}{3}$$
I am really confused. Is there something wrong with my steps?
 A: So $<BEC = \theta$ and thus $<BCE = 180^{\circ}-2\theta$
So $<ECD= 2\theta-90^{\circ} = <EDC$ and thus $<CED = 360^{\circ}-4\theta $
So $<DEA = 3\theta -180^{\circ} = < EAD$
So we have: $$\theta +(3\theta -180 ^{\circ}) = 90^{\circ} \implies \theta = 67,5^{\circ}$$ 
So $$\tan \theta = 1+\sqrt{2}$$
A: There is no such triangle.

By angle-chasing, $\theta=\frac{3}{8}\pi$, which yields $\tan(\theta)=1+\sqrt{2}$.

But then, since $\theta$ is uniquely determined, the diagram is already forced, up to similarity, so the ratio $AE{\,:\,}\!BE$ is uniquely determined, and can't just be arbitrarily specified as $5{\,:\,}2$ (unless it actually is $5{\,:\,}2$).

Given the known value of $\theta$, let's find the ratio $AE{\,:\,}\!BE$ . . .

Without loss of generality, we can assume $BC=1$.

Then by right-triangle trigonometry, we get
$$
AB=\text{sec}(\theta)
$$
and by the law of sines in triangle $BCE$, we get
$$
\frac{BE}{\sin(\pi-2\theta)}=\frac{1}{\sin(\theta)}$$
which yields
$$BE=\frac{\sin(\pi-2\theta)}{\sin(\theta)}=\frac{\sin(2\theta)}{\sin(\theta)}=2\cos(\theta)$$
hence
\begin{align*}
\frac{AE}{BE}&=\frac{AB-BE}{BE}\\[4pt]
&=\frac{AB}{BE}-1\\[4pt]
&=\frac{\text{sec}(\theta)}{2\cos(\theta)}-1\\[4pt]
&=\frac{\text{sec}^2(\theta)-2}{2}\\[4pt]
&=\frac{\tan^2(\theta)-1}{2}\\[4pt]
&=\frac{\left(1+\sqrt{2}\right)^2-1}{2}\\[4pt]
&=1+\sqrt{2}\\[4pt]
\end{align*}
hence $AE{\,:\,}\!BE\ne 5{\,:\,}2$.
