四边形ABCD,AB平行CD,AD平行BC,点E是AD的中点,点F是CD上一点,若△ABE面积=8,△DEF的面积=3,则△BEF=?

2024-12-01 18:22:10
推荐回答(1个)
回答1:

S△ABE = 8 , ,S△DEF = 3
S◇ABCD = 4S△ABE = 32
E是AD的中点 S△DCE = S△ABE = 8
S△CEF = S△CDE - S△DEF = 5
CF / DF = S△DCE / S△DEF = 5 / 3
CF / CD = 5 / 8
S△BCF = (5 / 16)S◇ABCD = 10
S△BEF = S◇ABCD - S△ABE - S△DEF - S△BCF = 11