A quadrilateral ABCD. The points BCD form an equilateral triangle. If AB=u; AD=v and Angle BAD= theta, prove that the area of the quadrilateral is

$\displaystyle \frac{1}{4}(\sqrt{3}u^2+\sqrt{3}v^2+2uv\sin \theta -2\sqrt{3}uv\cos \theta )$

thanks as always!