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.