Diagonals of a parallelogram bisect each other

Solution 1:

The conventions to using complex numbers to analytically analyze plane geometry problems are as follows.

First, pick a suitable origin -- in this case probably one of the vertices of your quadrilateral, say $A$ in $ABCD$. Then define complex numbers to represent all the "wiggly bits" of your diagram -- here you will want to normalize one edge of your parallelogram to be $1$, i.e. WLOG let $B = 1 \in \mathbb C$, and let $D = z \in \mathbb C$.

Now the next step is to compute whatever it is you want to find in 2 ways and verify that the complex expressions match. Here we want to compute the midpoints of $AC$ and $BD$ and verify that they are the same.

$C$ is located at $1+z$, so to find $AC$ we simply use the midpoint formula for complex numbers:
$z_M = \dfrac{z_P + z_Q}2$ iff $M$ is the midpoint of $PQ$.

There are plenty of these kinds of formulas to verify collinearity, concurrency, perpendicularity/parallelity of lines, cyclicity of quads, etc. They can be rederived or looked up.

For instance, in the second part, we specify that $ABCD$ is additionally a rhombus; this is equivalent to setting $|z| = 1$ or $z \bar z = 1$. To test that the diagonals are orthogonal, you can use the rule
$ i \dfrac{ z_P - z_Q }{z_R - z_S} \in \mathbb R$ iff $PQ \perp RS$, i.e.
$ \dfrac{z_P - z_Q }{ z_R - z_S} = -\overline { \left ( \dfrac{z_P- z_Q }{ z_R - z_S} \right )}. $