Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 12 / a / Solution

The relation rearranges to $(z_1+z_3)/2=(z_2+z_4)/2$, so the diagonals $AC$ and $BD$ have the same midpoint. Equivalently $z_2-z_1=z_3-z_4$ and $z_4-z_1=z_3-z_2$: each pair of opposite sides has the same displacement vector. Thus the given quadrilateral is a parallelogram. The identities also describe degenerate configurations if coincident or collinear vertices are allowed.