一道几何数学题。四边形abcd中角a等于60度。

DC=x,AB=yED=sqrt(3)*x,EC=2xAE=2y,EB=sqrt(3)*y带入EA-ED=DA,EB-EC=BC2y-sqrt(3)*x=4,sqrt(3)*y-2x=2解得x=4*sqrt(3)-4,y=8-2*sqrt(3)SABCD=sqrt(3)/2*(y²-x²)=6*sqrt(3)

解；延长DC交AB的延长线于点E,在RtΔADE和RtΔCBE中∵∠ADE=∠CBE ∠E=∠E∴ΔADE∽ΔCBE在RtΔADE中∵∠A=60^0∴∠E=30^0ctg30^0=ED/AD∴ED=4√3∴SΔADE=(1/2)×4×4√3=8√3∵SΔADE/SΔCBE=（AD/BC)^2=(4/2)^2=4即SΔCBE=(8√3)/4=2√3∴S四边形ABCD=SΔADE-SΔCBE=8√3-2√3=6√3