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.
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
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.