2011邢台一模:如图一,点E在AB上,点C在DB上,△ABC≌△DBE,AC与DE相交于点F,(1）求证∠AED=∠D

1
∵∠DCA=∠B+∠A
∠AED=∠B+∠D
又∵△ABC≌△DBE
∴∠A=∠D
∴∠DCA=∠AED
2;
∵△BCA≌△BED
∴ ∠BCA=∠BDE
∵∠ABC=∠ABE+∠EBC
∠EBD=∠EBC+∠DBC
∴∠ABE=∠CBD
∵BA=BD BE=BC
∴△ABE≌△BCD
∴∠BEA=∠BCD
∵∠BCF+∠BCD+∠ACD=360°
∠BEA+∠AED+∠BED=360°
∴∠AED=∠DCA
3
垂直关系
连结AD,BF并沿长BF与AD相交于H
∵△ABC≌△DBE
∠BAF=∠EAF+∠BAE
∠FDB=∠FDC+∠CDB
∠BAE=∠CDB
∠BAF=∠FDB
∴∠EAF=∠FDC
∵△ABE≌△BDC
∴CD=EA
∴△EAF≌△CDF
∴AF=FD
∵BA=BD AF=FD
∴BF⊥AD
好麻烦的题啊!