Prove that ABCD is a parallelogram

[Grand]

Homework Statement

Let $$z_1$$, $$z_2$$, $$z_3$$ and $$z_4$$ be te position vectors of the vertices of quadrilateral AMCD. Prove that ABCD is a parallelogram if and only if $$z_1-z_2-z_3+z_4=0$$.

Homework Equations

The Attempt at a Solution

The solution obviously uses the fact that collinear vectors of equal magnitude are equal, but I get $$z_1-z_2+z_3-z_4=0$$. Am I missing something obvious or is it just notations issue with the vertices of the quadrilateral.

 

[HallsofIvy]

$z_1- z_2$ is the side from "$z_2$" to "$z_1$" and $z_3- z_4$ is the side from "$z_4$" to "$z_3$". Since this is a parallelogram those sides are parallel and equal in length. In other words, the vectors are equal: $z_1- z_2= z_3- z_4$ whence $z_1- z_2- z_3+ z_4= 0$

 

[Grand]

That is true only if you haven't numbered the vertics clockwise, but in the form of a 'z' (if you get what I mean). In my picture, $$z_1$$ is adjacent to $$z_2$$ and $$z_4$$, and $$z_2$$ is adjacent to $$z_1$$ and $$z_3$$, so in my case the vector $$z_1-z_2$$ is opposite to the vector $$z_3-z_4$$.