For reference: Find the ratio between the segments $\frac{CE}{AE+DE}$
(Answer:$1$)
(The original figure)
My progress:
From geogebra I believe that the correct figure would have $AB$ and $BC$ tangent to the circumference and some important relationships of similarity and congruence I could not demonstrate.
$\triangle AEC \sim \triangle EDB \implies \frac{AE}{DE} = \frac{AC}{BD}=\frac{CE}{BE}\\ CE =\frac{AC.BE}{BD}=\frac{BE.AE}{DE}$
$AEDC$ is cyclic
But I couldn't find other relationships
Say, $\angle CEF = \theta~$ then $\angle AGB = 120^\circ - \theta$
Note that $\angle DBF = \angle DEF = 60^\circ - \theta$
That leads to $\angle BCA = 60^\circ$ and $ACDE$ is cyclic.
As chords $AC$ and $CD$ both make the same angle $(60^\circ)$ on the circle through $ACDE$, they must be equal. So we conclude that $\triangle ACD$ is equilateral.
$ \therefore AD = AC = CD$
Draw angle $60^\circ$ on $EA$ to intersect $CE$ at $H$.
We have $AH = EH = AE$ and $\angle AHC = 120^\circ$
As $\angle ACH = \angle ADE$, $\triangle ACH \cong \triangle ADE$
So we have, $CH = DE$ and $CE = CH + EH = AE + DE$