We use complex numbers with $(ACE)$ as the unit circle. The angle condition implies that $B$ is the intersection of tangents to $(ACE)$ at $A$ and $C$, so $b=\frac{2ac}{a+c}$. Since $DC=DE$, we can write $d=\lambda(c+e)$ for $\lambda\in\mathbb{R}$. Finally, $EC$ bisects $\angle AED$, so \[\frac{(d-e)(a-e)}{(c-e)^2}\in\mathbb{R}\implies \lambda=\frac{c^2+ae}{(c+a)(c+e)},d=\frac{c^2+ae}{c+a}.\]The midpoint of $BD$ is $\frac{c^2+2ac+ae}{2(a+c)}$. We finish by observing \[\frac{\frac{c^2+2ac+ae}{2(a+c)}-e}{c-e}=\frac{c^2+2ac-2ce-ae}{2(a+c)(c-e)}\in\mathbb{R}.\]

Cool problem!

It's easy to solve with the law of sinuses and similarity of two isosceles triangles