Prove that angles are equal

Question

Let $\triangle ABC$ be a triangle. Point $D$ is on $BC$ edge and $E$ on $AB$. We know that $BD=AC$, $AD=AE$, $AB^2=AC*BC$. Prove, that $\angle BAD$ $=$ $\angle CEA$ . I think, that Menelaus' theorem can be helpful there, but I don't know how to solve it. Sorry for my English .

Answer 1

Note that $ab=c^2$ implies that $$BD \cdot BC=BA^2$$ and hence $$\frac{BA}{BD}=\frac{BC}{BA}$$ and hence it's easy to see that the triangles $BAD$ and $BCA$ are similar since they share a common angle at $B$ and have the same side ratio.

Thus, we conclude that $\measuredangle EAC=\measuredangle ADB$.

Now, we note that the triangles $BAD$ and $CEA$ must be congruent by the above angle equality and $AE=AD$ as well as $BD=AC$.

And finally this implies that we must also have $\measuredangle CEA=\measuredangle BAD$ as desired.