Let $K$ be the intersection of diagonals of cyclic quadrilateral $ABCD$, where $|AB|=|BC|$, $|BK|=b$, and $|DK|=d$. What is $|AB|$?

$ 
\textbf{(A)}\ \sqrt{d^2 + bd}
\qquad\textbf{(B)}\ \sqrt{b^2+bd}
\qquad\textbf{(C)}\ \sqrt{2bd}
\qquad\textbf{(D)}\ \sqrt{2(b^2+d^2-bd)}
\qquad\textbf{(E)}\ \sqrt{bd}
$
1. **Identify the given information and the goal:**
   - We have a cyclic quadrilateral \(ABCD\) with diagonals intersecting at point \(K\).
   - Given: \(|AB| = |BC|\), \(|BK| = b\), and \(|DK| = d\).
   - We need to find the length of \(|AB|\).

2. **Use properties of cyclic quadrilaterals:**
   - In a cyclic quadrilateral, opposite angles are supplementary.
   - The diagonals of a cyclic quadrilateral intersect at an angle such that the opposite angles are equal.

3. **Apply the given conditions:**
   - Since \(|AB| = |BC|\), triangle \(ABK\) is isosceles with \(\angle BAK = \angle BCK\).
   - Similarly, triangle \(BDK\) is isosceles with \(\angle BDK = \angle BCK\).

4. **Use the similarity of triangles:**
   - Since \(\angle BAK = \angle BDK\) and \(\angle BKA = \angle BKD\), triangles \(\triangle BAK\) and \(\triangle BDK\) are similar by AA similarity criterion.
   - Therefore, \(\triangle BAK \sim \triangle BDK\).

5. **Set up the proportion from the similarity:**
   \[
   \frac{AB}{BK} = \frac{BD}{AB}
   \]
   - Let \(AB = x\). Then, substituting the given values, we get:
   \[
   \frac{x}{b} = \frac{b + d}{x}
   \]

6. **Solve the proportion for \(x\):**
   \[
   x^2 = b(b + d)
   \]
   \[
   x = \sqrt{b(b + d)}
   \]

7. **Substitute back to find \(|AB|\):**
   \[
   |AB| = \sqrt{b^2 + bd}
   \]

Conclusion:
\[
\boxed{\sqrt{b^2 + bd}}
\]