As shown in the figure, in the parallelogram ABCD, AB = 3, AD = 4, ∠ ABC = 60, the midpoint E passing through BC is EF⊥AB, and the vertical foot is point F, DC.

In the right triangle BEF, because ∠ ABC = 60, BE=BC/2=CD/2=2.

So BF=BE/2= 1,

So AF=AB-BF=3- 1=2,

So △ADF area =2√3

You can also prove that △ BEF △ CEH.

So S△BEF=S△CEH, EF=CE.

So △DEF and △DEH are triangles with the same base and height.

So S△DEF=S△DEH

So S△DEF=( 1/2) (parallelogram ABCD area -△ADF area)

Because the area of parallelogram ABCD = AB * CD * (√ 3/2) = 6 √ 3.

So s △ def = (1/2) * (6 √ 3-2 √ 3) = 2 √ 3.