Find the angle;}

Let's call $\angle CDE$ $x$ . Because $\triangle ABD$ lies on circle $B$ and it's centerpoint, angles $\angle ADB$ and $\angle BAD$ are equal, and therefore $\angle ABD = 180 - 2x$ . Because $\angle ABD + \angle ABE = 180^{\circ}$ , $\angle ABE$ has to be $2x$ . Since $\triangle EAB$ is an iscoceles triangle, because $E$ and $B$ lie on circle $A$ , and $A$ is that circle's centerpoint. Therefore $\angle AEB = 2x$ . Also $\triangle CAE$ is an isosceles triangle, so $\angle CEA = 63^{\circ}$ . Now we can determine the value of $x$ using that the sum of the three angles of $\triangle CED$ have to add up to $180^{\circ}$ :

$63^{\circ} + 63^{\circ} + x + 2x = 180^{\circ}$

$3x = 54^{\circ}$

$x = 18^{\circ}$

And therefore $\angle CDE = 18^{\circ}$ .