Given points P(0, 3), Q(4, 1), R(2, -3), S(-2, -1), show PR = QS, find the midpoint of QS (N), and prove PN is perpendicular to QS.

For PR, use $P(0, 3)$ and $R(2, -3)$:

For QS, use $Q(4, 1)$ and $S(-2, -1)$:

Since $PR = \sqrt{40}$ and $QS = \sqrt{40}$, we have $PR = QS$.

For N, the midpoint of QS, use $Q(4, 1)$ and $S(-2, -1)$:

The coordinates of N are $\boxed{(1, 0)}$.

For PN, use $P(0, 3)$ and $N(1, 0)$:

For QS, use $Q(4, 1)$ and $S(-2, -1)$:

Since the product of their gradients is -1, PN is perpendicular to QS.

$m_{PN} = -3$.

For NR, use $N(1, 0)$ and $R(2, -3)$:

Since $m_{PN} = m_{NR} = -3$ and N is a common point, P, N, and R are collinear.

From 1.2, N is the midpoint of QS, and $N = (1, 0)$.
Let's find the midpoint of PR using $P(0, 3)$ and $R(2, -3)$:

Since the midpoint of PR is also $(1, 0)$, N is the common midpoint of both diagonals PR and QS. This means the diagonals bisect each other, so PQRS is a parallelogram.

From 1.1, we showed that $PR = QS = \sqrt{40}$.
A parallelogram with equal diagonals is a rectangle.

From 1.3, we showed that $m_{PN} \times m_{QS} = -1$. Since N lies on PR (from 1.4), this implies that the diagonal PR is perpendicular to the diagonal QS.
A parallelogram with perpendicular diagonals is a rhombus.

Since PQRS is both a rectangle (diagonals are equal) and a rhombus (diagonals are perpendicular), it must be a square.

The quadrilateral PQRS is a square.
Reason: Its diagonals bisect each other (making it a parallelogram), are equal in length (making it a rectangle), and are perpendicular (making it a rhombus). A quadrilateral with all these properties is a square.