Question

如图，已知四边形ABCD是菱形，F是AB上一点，DF交AC于E，连接BE，求证：角AFD=角CBE

Answer 1

∵四边形ABCD是菱形
∴∠ACB=∠ACD,BC=CD,EC=EC
∴⊿BCE≌⊿DCE
∴∠CDE＝∠CBE
∵AB∥CD
∴∠AFD＝∠CDE
∴∠CDE＝∠CBE

Answer 2

设菱形对角线AC和BD交于O
∵ABCD是菱形
∴AC⊥BD
OD=OB
∠ABD=∠DBC=∠ADB=∠BDC
∠DAB+∠ADC=180°
∠ADC=2∠ADC=2∠BDC
∵E在AC上
∴DE=BE
∴∠EDB=∠EBD
∵∠CBE=∠DBC+∠EBD=∠ADB+∠EDB
∠AFD=180°-∠DAB-∠ADF
=180°-(180°-∠ADC)-(∠ADB-∠EDB)
=∠ADC-∠ADB+∠EDB
=2∠ADB-∠ADB+∠EDB
=∠ADB+∠EDB
∴∠AFD=∠CBE

Answer 3

由菱形的对称性知∠CDE＝∠CBE
∵AB∥CD
∴∠AFD＝∠CDE
∴∠CDE＝∠CBE

Answer 4

请问AB∥CD 哪里来的

Answer 5