Question

如图,已知平行四边形ABCD,AC与BD交于点O,BE⊥AD交DA延长线于点E,BF⊥CD交DC延长线于F

Answer 1

∠EAB=∠BCF=∠D=∠ABC

∴∠EBA=∠CBF

又∵∠EBF=2∠ABC

∴∠EBA+∠CBF=∠ABC

∴∠EBA=∠CBF=1/2∠ABC=1/2∠BCF=1/2∠EAB

即∠EBA=1/2∠EAB

∠CBF=1/2∠BCF

∴∠EBA=∠CBF=30°

∴AB=CD=2AE=8

BF=根3CF=6根3

∴S=CDxBF=48根3