Referred to an origin O, the position vectors of four non-collinear points A, B, C and D are a, b, c and d respectively. Given that $a - b = d - c$, show that ABCD is a parallelogram.

1. Express the given vector equation in terms of displacement vectors.

$\vec{OA} = \mathbf{a}, \quad \vec{OB} = \mathbf{b}, \quad \vec{OC} = \mathbf{c}, \quad \vec{OD} = \mathbf{d}$
$\mathbf{a} - \mathbf{b} = \vec{OA} - \vec{OB} = \vec{BA}$
$\mathbf{d} - \mathbf{c} = \vec{OD} - \vec{OC} = \vec{CD}$
Given $\mathbf{a} - \mathbf{b} = \mathbf{d} - \mathbf{c}$, we have:
$\vec{BA} = \vec{CD}$

2. Interpret the vector equality geometrically to deduce the shape of the quadrilateral.

The vector equality $\vec{BA} = \vec{CD}$ indicates that the line segment $BA$ is parallel to the line segment $CD$ (same direction) and they are equal in length (same magnitude).
Because the points $A, B, C,$ and $D$ are non-collinear, having one pair of opposite sides that are both equal and parallel is sufficient to conclude that quadrilateral $ABCD$ is a parallelogram.

1. Interpret the given vector magnitudes as the lengths of the diagonals of the parallelogram.

The vector $\mathbf{a} - \mathbf{c} = \vec{OA} - \vec{OC} = \vec{CA}$, so $|\mathbf{a} - \mathbf{c}|$ represents the length of the diagonal $AC$.
Similarly, $\mathbf{b} - \mathbf{d} = \vec{OB} - \vec{OD} = \vec{DB}$, so $|\mathbf{b} - \mathbf{d}|$ represents the length of the diagonal $BD$.

2. Identify the specific shape of the parallelogram using the properties of its diagonals.

Given $|\mathbf{a} - \mathbf{c}| = |\mathbf{b} - \mathbf{d}|$, it follows that the diagonals $AC$ and $BD$ are equal in length.
A parallelogram with equal diagonals must be a rectangle.