How to prove that it is possible to make rhombuses with any number of interior points? I was given some square dot paper which can be found on this link: http://lrt.ednet.ns.ca/PD/BLM/pdf_files/dot_paper/sq_dot_1cm.pdf  and was told to draw a few rhombuses with the vertices on the dots but no other dots on the edges.
Here is an example:
This is not a rhombus it just shows how the shape must be lay out. 

                   A = 6

The number of interior dots would be represented by A. I was then told that it is possible to draw rhombuses with any number of interior points. All I had to do was prove this.
I started by trying to draw these shapes and came up with some of the below:

A = 2

A = 4

A = 9  
However, I am still unsure of how you can prove something like this since there may be numerous ways of drawing these rhombuses.
I would appreciate support.
Thank you :) 

The proof is straight-forward. Lets use the property that rhombus is a quadrilateral whose diagonals are perpendicular bisectors. We'll accommodate all the interior points we want along a line. 


*

*Odd number of interior points


Say we want $p_o$ interior points; consider $p_o$ consecutive points along a line (notice the square paper has two symmetries - straight and diagonal, so we can choose either a straight or a diagonal line. Lets choose straight line). Now take one point each on either side of these $p_o$ points; they shall serve as two of the vertices of our desired rhombus. Consistent with our definition, choose the point which bisects the distance between these $p_o+2$ points; from the point of bisection, draw a perpendicular line. Immediate points on each side that this line intersects are the other two vertices of our rhombus. Join the four vertices and we have our desired rhombus. 


*Even number of interior points


We follow the same procedure, but since the line perpendicularly bisecting the distance between two $p_e+2$ points will never intersect an other point on the square paper, we make use of its diagonal symmetry. So, we choose our $p_e$ points along a diagonal instead and follow the same procedure.
So, the proof follows from Rhombus's diagonals being perpendicular bisectors.
