Proof for parallelogram law of vector addition The Statement of Parallelogram law of vector addition is,

If two vectors are considered to be the adjacent sides of a parallelogram, then the resultant of two vectors is given by the vector that is a diagonal passing through the point of contact of two vectors.

But how do we justify that the resultant is along the diagonal? Is it based on experimental evidence, or is it something that can be proved? 
I know, the question might sound pretty obvious, but I'm new to this stuff. :)

Addition of Vectors basically found their origin from the Triangular Law of Vector Addition


The triangle law of vectors basically is a process that allows one to take two vectors, draw them proportional to each other, connect them head to tail, then draw the resultant vector as a result of the third side that is missing.

The Parallelogram law is just a furthermore explanation of Triangular law,

If two vectors are considered to be the adjacent sides of a Parallelogram, then the resultant of two vectors is given by the vector which is a diagonal passing through the point of contact of two vectors.

The proof for the resultant vector in Parallelogram addition is as follows,
Consider a parallelogram $OABC$ as shown in the figure,

Let $P \;\&\; Q$ be two adjacent sides of parallelogram, and $R$ be the resultant vector obtained by addition of vectors $P \;\&\; Q$,
Now, drop a perpendicular from $C$ on $OA$ so that they meet at $A$.
From right angled triangle $\Delta OCD \;,\; OC^2=OD^2+DC^2$
$$[OD=OA+AD]$$
$$R^2=(OA+AD)^2+DC^2$$
$$R^2=OA^2+AD^2+2OA.AD+DC^2$$
From $\Delta$ADC, $AC^2=AD^2+DC^2$
and also $\cos\theta= \frac{AD}{AC}$
And hence $R^2= OA^2+AC^2+2 OA.AC \cos\theta$
And substituting A and B $$R^2=A^2+B^2+2 A.B \cos\theta$$
