Why am I getting two different answer for this geometry question?

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 $$\tan \theta = 1+\sqrt{2}$$

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 $AE{\,:\,}\!BE\ne 5{\,:\,}2$.