jmc

Both angles $\angle BAD$ and $\angle CBE$ subtend arc $BE$, so $\angle CBE=\angle BAE=43^{\circ }$. Triangle $BCE$ is isosceles with $BC=CE$, since these are tangent from the same point to the same circle, so $\angle CEB=\angle CBE=43^{\circ }$.

Finally, $\angle AEB=90^{\circ }$ since $AB$ is a diameter, so $\angle BED=90^{\circ }$. Therefore, $\angle CED=\angle BED-\angle BEC=90^{\circ }-43^{\circ }=\boxed{47^{\circ }}$.