cosine formula

Suppose ABCD is a parallelogram with AB//CD and BC//AD

Also, let ∠ABC = θ, then ∠BAD = 180 - θ

Applying the cosing formula for △ABC and △BAD:

AD2 = AB2 + BC2 - 2 AB BC cos θ

BC2 = BA2 + AD2 - 2 BA AD cos (180 - θ)

= CD2 + AD2 + 2 BA BC cos θ

AD2 + BC2 = AB2 + BC2 + CD2 + AD2

Let the two sides of the parallelogram is a and b with the angle θ (where θ is an obtuse angle)Then the square of the length of the long diagonal= a^2 + b^2 - 2ab cosθThe square of the length of the shorter diagonal= a^2 + b^2 - 2ab cos(180 - θ)= a^2 + b^2 + 2ab cosθSo, the sum of the squares of the lengths of two diagonals = a^2 + b^2 + a^2 + b^2= sum of the squares of the lengths of the four sides