Geometry and van Aubel's theorem

Question

Given an arbitrary planar quadrilateral, place a square outwardly on each side, and connect the centers of opposite squares. Then van Aubel's theorem states that the two lines are of equal length and cross at a right angle. How can I prove that the area $K$ of the larger quadrilateral is $K= S + \frac {x^2 + y^2}{4}$, where $S$ is the area of the quadrilateral and $x$, $y$ are the lengths of its diagonals

Answer 1

You may assume that the vertices $a_k$ $(k\ {\rm mod}\ 4)$ of the given (convex) quadrilateral $Q_a$ are numbered counterclockwise. Consider the $a_k$ as complex numbers. Then the vertices $b_k$ of the larger quadrilateral $Q_b$ are given by $$b_k={a_k+a_{k-1}\over 2}-i{a_k-a_{k-1}\over 2}={1-i\over2} a_k+{1+i\over2}a_{k-1}\ .$$ In order to compute the areas of $Q_a$ and $Q_b$ use the shoelace formula, translated into this complex setting.

Using complex numbers for this problem has the following advantage over standard vector algebra: Rotating by $90^\circ$ in counterclockwise direction is just multiplication by $i$, hence comes for free.