hint or idea

Answer

The answer is $40$. Indeed, if the center of the circle is $O$, then $AO$, $BO$, $CO$, $DO$, $EO$ are angle bisectors, so $\angle OAE = \angle OEA = 50^{\circ}$ and $AO = OE$. Furthermore, $\triangle DOE \cong \triangle DOC$ since they have a common side and two equal angles. Thus $OC = OE$, so $O$ is the circumcenter of $ACE$ and $\angle ACE = \frac{\angle AOE}{2} = 40^{\circ}$.

Let $\angle BAC= \alpha, \angle CED = \beta$. $\angle BCD$ is the inscribed angle of $BD$, $\alpha$ is the inscribed angle of $BC$, and $\beta$ is the inscribed angle of $CD$. Thus $100+\alpha+\beta=180$ i.e. $\alpha+\beta=80$. Here $\angle ACE = 180-\angle CAE -\angle CEA = 180-(200-\alpha-\beta)=60$.

(bad proof style but why not) $\angle ACE = 100 - (\angle BCA + \angle DCE) = 100 - (\angle EAD + \angle AEB) = 100 - (\angle A - \angle DAB + \angle E - \angle BED) = 100 - (100 - 80 + 100 - 80) = 60$. QED.

@above the quadrilateral *contains* a circle and is not *contained in* a circle!

Has everyone made the same mistake that the pentagon is inside the circle?