Notes
four isosceles triangles in a circle solution

Four Isosceles Triangles in a Circle

Solution by Isosceles Triangles and Cyclic Quadrilateral

Let $a$ be angle $C \hat{A} E$. As triangle $E A B$ is isosceles, angle $A \hat{E} B$ is $180^\circ - 2 a$. As triangle $E C A$ is also isosceles, angle $E \hat{C} A$ is also $180^\circ - 2 a$ and angle $C \hat{E} A$ is also $a$. Since triangle $C \hat{E} B$ is isosceles, angle $C \hat{E} B$ is also $180^\circ - 2 a$. Looking at angle $C \hat{E} A$, then $a = 2 (180^\circ - 2 a)$ which means that $a = 72^\circ$. Lastly, the quadrilateral $A C D E$ is cyclic so angle $E \hat{D} C$ is $180^\circ - 72^\circ = 108^\circ$.