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+CE^2`

`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)`

`AE =AB*cos(/_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**.