For $z_1, z_2 \in \mathbb C$ prove that $|z_1+z_2|^2+|z_1-z_2|^2=2(|z_1|^2+|z_2|^2)$, and interpret this result geometrically. 
For $z_1, z_2 \in \mathbb C$ prove that $|z_1+z_2|^2+|z_1-z_2|^2=2(|z_1|^2+|z_2|^2)$, and interpret this result geometrically.

We have
\begin{align*}
 |z_1+z_2|^2+|z_1-z_2|^2&=(z_1+z_2)(\overline{z_1+z_2})+(z_1-z_2)(\overline{z_1-z_2})\\
 &=(z_1+z_2)(\overline{z_1}+\overline{z_2})+(z_1-z_2)(\overline{z_1}-\overline{z_2})\\
 &=z_1\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_2}+\overline{z_1}z_2+z_1\overline{z_1}+z_2\overline{z_2}-z_1\overline{z_2}-\overline{z_1}z_2\\
 &=2z_1\overline{z_1}+2z_2\overline{z_2}\\
&=2|z_1|^2+2|z_2|^2\\
 &=2(|z_1|^2+|z_2|^2).
\end{align*}
However, I am having trouble "interpreting this result geometrically"?
What does this mean?
 A: User Al Jebr has proven the parallelogram law:
https://en.wikipedia.org/wiki/Parallelogram_law
(Robert Z).
Consider the complex plane:
Draw the vectors $\vec z_1$, $\vec z_2$, not collinear from the origin.
Let $\vec OA:= \vec z_1$; $\vec OC := \vec z_2$.
Complete the parallelogram  $OABC$ by drawing a parallel through $A$, parallel to $\vec z_2$, and a parallel through C, parallel to $\vec z_1$.
These $2$ parallels intersect at point $B$.
$OABC$ is a parallelogram.
$|\vec z_1+\vec z_2| = \overline{OB}=$
length of diagonal $OB$.
$|\vec z_1- \vec z_2| =\overline{OC}=$ 
length of diagonal $OC$.
$|\vec z_1| = \overline{OA} =$  length of side $OA$.
$|\vec z_2| = \overline{OC} =$ length of side $OC$.
Parallelogram law:
$[\overline{OB}]^2 +[\overline{AC}]^2 = $
$2(\overline{OA})^2 +2(\overline{OC})^2$.
A: in components:
$$
(x_1+x_2)^2 +(y_1+y_2)^2 + (x_1-x_2)^2 +(y_1-y_2)^2 = 2(x_1^2 + y_1^2 + x_2^2 + y_2^2)
$$
if a parallelogram has sides $a,b$ making an angle of $\theta$ then the cosine rule tells us that if the two diagonals have lengths $d_1, d_2$ then
$$
d_1^2 = a^2 +b^2 - 2ab\cos\, \theta \\
d_2^2 = a^2 +b^2 + 2ab\cos\, \theta
$$
now add the two
