Geometry of triangles, circumcircles and incircles

Question

Let $\triangle ABC$ be a triangle with incenter $I$. The angle bisector of $\angle BCA$ meets the line $\overline{AB}$ at $D$ and the circumcircle of $\triangle ABC$ at $C$ (obviously) and $E$. For some numeric values is that the equation $\overline{EI}^2= \overline{EC}\ \overline{ED}$ seems to hold for every position of $A$, $B$, and $C$. I'd be grateful for every hint as I have absolutely no idea how to prove that.

Answer 1

Yes, you are right!

We see that: $$\Delta EBC\sim\Delta EDB$$ and from here $$\frac{EB}{ED}=\frac{EC}{EB},$$ which gives $$EB^2=EC\cdot ED$$ and since $$\measuredangle EBI=\frac{\beta+\gamma}{2}=\measuredangle EIB,$$ we obtain $$EB=EI$$ and we are done!