Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:
$ \textbf{(A)}\ \text{a square} \qquad\textbf{(B)}\ \text{a rectangle} \qquad\textbf{(C)}\ \text{a parallelogram with unequal sides} \\
\textbf{(D)}\ \text{a rhombus} \qquad\textbf{(E)}\ \text{a quadrilateral with no special properties}$
1. Consider a rectangle \(ABCD\) with vertices \(A\), \(B\), \(C\), and \(D\). Each angle of the rectangle is \(90^\circ\).
2. Trisect each angle of the rectangle. This means each \(90^\circ\) angle is divided into three \(30^\circ\) angles.
3. Let the trisectors of \(\angle A\) meet the trisectors of \(\angle B\) at point \(P\), the trisectors of \(\angle B\) meet the trisectors of \(\angle C\) at point \(Q\), the trisectors of \(\angle C\) meet the trisectors of \(\angle D\) at point \(R\), and the trisectors of \(\angle D\) meet the trisectors of \(\angle A\) at point \(S\).
4. We need to determine the shape formed by the points \(P\), \(Q\), \(R\), and \(S\).

To analyze the shape, consider the following properties:
- The diagonals of the quadrilateral \(PQRS\) lie along the lines joining the midpoints of the opposite sides of the rectangle.
- These diagonals are perpendicular to each other because they are formed by the trisectors of the \(90^\circ\) angles, which are \(30^\circ\) and \(60^\circ\) lines.
- The diagonals are of different lengths because the rectangle's sides are of different lengths.

Since the diagonals are perpendicular bisectors of each other and of different lengths, the quadrilateral \(PQRS\) must be a rhombus.

\(\blacksquare\)

The final answer is \(\boxed{\textbf{(D)}}\).