I need to prove parallelogram law using the law of cosines.

2AB^2 + 2BC^2 = AC^2 + BD^2
Please, help me to do this.

4 answers

Let
θ = angle DAB
Ø = angle ABC

By the law of cosines,

BD^2 = AD^2 + AB^2 - 2 AD AB cosθ
AC^2 = BC^2 + AB^2 - 2 BC AB cosØ

Now, since AD = BC,

BD^2 + AC^2 = 2AD^2 + 2AC^2 - 2 AD AB (cosθ + cosØ)

Since θ+Ø = 180°
cosθ = -cosØ

'nuff said.
Oops. In the long line, on the right side, AC should read AB
ehh, still don't get it.
where do you get stuck? I'll try to clarify. Draw a diagram to follow along.