$ ABCD$ is a cyclic quadrilateral. If $ \angle B \equal{} \angle D$, $ AC\bigcap BD \equal{} \left\{E\right\}$, $ \angle BCD \equal{} 150{}^\circ$, $ \left|BE\right| \equal{} x$, $ \left|AC\right| \equal{} z$, then find $ \left|ED\right|$ in terms of $ x$ and $ z$.

$\textbf{(A)}\  \frac {z \minus{} x}{\sqrt {3} } \qquad\textbf{(B)}\  \frac {z \minus{} 2x}{3} \qquad\textbf{(C)}\  \frac {z \plus{} x}{\sqrt {3} } \qquad\textbf{(D)}\  \frac {z \minus{} 2x}{2} \qquad\textbf{(E)}\  \frac {2z \minus{} 3x}{2}$
1. Given that \(ABCD\) is a cyclic quadrilateral, we know that opposite angles sum to \(180^\circ\). Since \(\angle B = \angle D\), we have \(\angle A + \angle C = 180^\circ\).

2. We are given \(\angle BCD = 150^\circ\). Let \(\angle BAC = \phi\). Since \(\angle B = \angle D\), we also have \(\angle BDC = \phi\).

3. By the Law of Sines in \(\triangle BCD\):
   \[
   \frac{BC}{\sin \angle BDC} = \frac{BD}{\sin \angle BCD}
   \]
   Substituting \(\angle BDC = \phi\) and \(\angle BCD = 150^\circ\):
   \[
   \frac{BC}{\sin \phi} = \frac{BD}{\sin 150^\circ}
   \]
   Since \(\sin 150^\circ = \sin 30^\circ = \frac{1}{2}\):
   \[
   \frac{BC}{\sin \phi} = \frac{BD}{\frac{1}{2}} \implies BC = BD \cdot \frac{\sin \phi}{\frac{1}{2}} = 2BD \cdot \sin \phi
   \]

4. From \(\triangle ABC\), using the Law of Sines:
   \[
   \frac{BC}{\sin \angle BAC} = \frac{AC}{\sin \angle ABC}
   \]
   Substituting \(\angle BAC = \phi\) and \(\angle ABC = 180^\circ - \phi\):
   \[
   \frac{BC}{\sin \phi} = \frac{z}{\sin (180^\circ - \phi)} = \frac{z}{\sin \phi}
   \]
   Therefore:
   \[
   BC = z
   \]

5. From step 3, we have \(BC = 2BD \cdot \sin \phi\). Equating this with \(BC = z\):
   \[
   z = 2BD \cdot \sin \phi
   \]

6. We need to find \(ED\) in terms of \(x\) and \(z\). Note that \(BD = BE + ED\). Let \(BE = x\), then:
   \[
   BD = x + ED
   \]

7. Substituting \(BD = x + ED\) into the equation \(z = 2BD \cdot \sin \phi\):
   \[
   z = 2(x + ED) \cdot \sin \phi
   \]

8. Solving for \(ED\):
   \[
   z = 2x \cdot \sin \phi + 2ED \cdot \sin \phi
   \]
   \[
   z - 2x \cdot \sin \phi = 2ED \cdot \sin \phi
   \]
   \[
   ED = \frac{z - 2x \cdot \sin \phi}{2 \cdot \sin \phi}
   \]

9. Since \(\sin \phi\) cancels out:
   \[
   ED = \frac{z - 2x}{2}
   \]

Conclusion:
\[
\boxed{\frac{z - 2x}{2}}
\]