题目内容

如图,在直角梯形ABCD中,ADBCABADBCCDBECD,垂足为E,点FBD上,连接AF、EF

(1)求证:DADE

(2)如果AFCD,求证:四边形ADEF是菱形.

证明:(1)∵ADBC,∴∠DBC=∠ADB

又∵BC=CD,∴∠DBC=∠BDC. 

∴∠ADB=∠BDC.······························ 1分

又∵∠ADB=∠BDC,BAADBECD,∴BA=BE

在RT△ABD和RT△EB中, BD=BDAB=BE

∴△ABD≌△EBD.   ···························· 2分

 ∴AD=ED.································· 3分

(2) ∵AFCD,∴∠BDC=∠AFD

又∵∠ADB=∠BDC,∴∠AFD=∠ADB. ∴AD=AF

又∵AD=DE,∴AF= DEAFCD.∴四边形ADEF为平行四边形.········· 6分

AD=DE ,∴四边形ADEF为菱形. ······················ 7分

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