The figure is attached below.
Suppose you have the parallelogram ABCD and a diagonal of it BC. In triangle ABC, you draw the height `BE_|_AC` .
Since triangle BCE has a 90 degree angle one can write
`BC^2 =BE^2 +(AC-AE)^2` (1)
Now from triangle ABE which also has a 90 degree angle one can express BE and AE as a function of `/_A` .
`BE = AB*sin(/_A)`
Which replaced in expression (1) above gives
`BC^2 =AB^2*sin^2(/_A) +[AC-AB*cos(/_A)]^2`
`BC^2 =AB^2*sin^2(/_A) +AC^2 -2*AC*AB*cos(/_A) +AB^2*cos^2(/_A)`
Finally because `sin^2(/_A) +cos^2(/_A) =1` one obtains
`BC^2 =AB^2 +AC^2 -2AB*AC*cos(/_A)`
Therefore the law of cosines was demonstrated using a parallelogram.
thanks alot (^_^)