Let $ ABCD$ be a quadrilateral with $ AB$ extended to $ E$ so that $ \overline{AB} \equal{} \overline{BE}$.  Lines $ AC$ and $ CE$ are drawn to form angle $ ACE$.  For this angle to be a right angle it is necessary that quadrilateral $ ABCD$ have:

$ \textbf{(A)}\ \text{all angles equal}$
$ \textbf{(B)}\ \text{all sides equal}$
$ \textbf{(C)}\ \text{two pairs of equal sides}$
$ \textbf{(D)}\ \text{one pair of equal sides}$
$ \textbf{(E)}\ \text{one pair of equal angles}$
1. Let \( AE \) be the diameter of a circle with center at \( B \). Since \( AB = BE \), \( E \) lies on the circle with diameter \( AE \).
2. For \(\angle ACE\) to be a right angle, point \( C \) must lie on the circle with diameter \( AE \). This is because the angle subtended by a diameter in a circle is a right angle (Thales' theorem).
3. Therefore, \( C \) must be equidistant from \( A \) and \( E \), implying that \( CB = AB = BE \).
4. The only way to ensure that \( CB = AB = BE \) is if all sides of the quadrilateral \( ABCD \) are equal. This would make \( ABCD \) a rhombus, where all sides are equal.

Thus, the necessary condition for \(\angle ACE\) to be a right angle is that all sides of the quadrilateral \( ABCD \) are equal.

The final answer is \(\boxed{B}\)