如图,已知在四边形ABCD中,AD∥BC,AD>BC,AB=DC,EA=ED,EB,EC分别交AD于点F,G 求证四边形FBGC是等腰梯形

∵AD∥BC,BE与CE相交于E,即BE、CE不平行,∴四边形FBCG是梯形,∵AD∥BC,AD>BC,∴四边形ABCD是梯形, ∵AB=CD,∴梯形ABCD是等腰梯形,∴∠BAD=∠CDA,∵AE=DE,∴∠EAD=∠EDA,∴∠EAD+∠BAD=∠EDA+∠CDA,即∠EAB=∠EDC,∴...