题目内容
如图,在直角梯形ABCD中,AD∥BC,AB⊥AD,BC=CD,BE⊥CD,垂足为E,点F在BD上,连接AF、EF.
(1)求证:DA=DE;
(2)如果AF∥CD,求证:四边形ADEF是菱形.
![]()
![]()
证明:(1)∵AD∥BC,∴∠DBC=∠ADB.
又∵BC=CD,∴∠DBC=∠BDC.
∴∠ADB=∠BDC.······························ 1分
又∵∠ADB=∠BDC,BA⊥AD,BE⊥CD,∴BA=BE.
在RT△ABD和RT△EB中, BD=BD, AB=BE.
∴△ABD≌△EBD. ···························· 2分
∴AD=ED.································· 3分
(2) ∵AF∥CD,∴∠BDC=∠AFD.
又∵∠ADB=∠BDC,∴∠AFD=∠ADB. ∴AD=AF.
又∵AD=DE,∴AF= DE且AF∥CD.∴四边形ADEF为平行四边形.········· 6分
∵AD=DE ,∴四边形ADEF为菱形. ······················ 7分
练习册系列答案
相关题目