Quadrilateral, Diagonal, and Circle

Since diagonal $AB$ is the diameter of the circle, triangles $ADB$ and $ACB$ are right triangles. Then

$\sin 45 = \dfrac{BD}{4}$ $\implies$ $BD=4 \sin 45 = 4\left(\dfrac{\sqrt{2}}{2}\right) = 2\sqrt{2}$

It follows that $AD=BD=2\sqrt{2}$ since triangle $ADB$ is an isosceles right triangle with $\angle ABD=45^\circ$ .

Applying pythagorean theorem on triangle $ACB$ , we get, $BC=\sqrt{4^2-2^2}=\sqrt{12}=2\sqrt{3}$

Applying Ptolemy's Theorem , we have

$(AB)(CD)=(AD)(CB)+(AC)(BD)$

$4(CD)=2\sqrt{2}(2\sqrt{3})+2(2\sqrt{2})$

$CD=\dfrac{4(\sqrt{6}+\sqrt{2})}{4}=\boxed{\sqrt{6}+\sqrt{2}}$