We have a triangle $\Delta ABC$. Point $D$ is on the $BC, E$ on $AB$, $BD=AC$, $AD=AE$ and ${AB}^2=AC\times BC$. Prove that $\angle BAD$ is equal to $\angle CEA$.
That's the theory.
I found that all I need to do is to prove a similarity of triangles $\Delta ABD$ and $\Delta AEC$ by proving that $AB$ is corresponding/equal to $CE$. I tried using Stewart's theorem for $\Delta ABC$ once $BC$ being the base and once $AB$ being the base, then solving them as a system of linear equations (substituting values that are the same), but nothing works.
Any help would be appreciated.
For starters, let we set $AC=u^2$ and $CB=v^2$. That implies $AB=uv$ and $$\cos\widehat{ACB} = \frac{u^4+v^4-u^2v^2}{2u^2 v^2}.$$ Now we just need to apply many times the cosine theorem. Since we have the cosine of $\widehat{ACB}$, we may compute $AD^2$ from the previous line and the identity $AC^2=CD^2=u^4$. That gives $AE^2$ in terms of $u,v$, too. If we compute $\cos\widehat{BAC}$, we may compute $CE^2$ in a similar way, since we know $AC^2$ and $AE^2$. That gives $\cos\widehat{CEA}$, in terms of $u$ and $v$, as a by-product. We know the squared lengths of the sides of $BAD$, hence we know $\cos\widehat{BAD}$, too, and we just need to compare it with the previously computed $\cos\widehat{CEA}$.
If the claim is true, this very brutal approach has to work.
You just have to be brave enough to perform the above computations for real.