已知:AB=AD,AC=AE,∠BAD=∠CAE=90°BF=EF求证:AF⊥CD

已知:AB=AD,AC=AE,∠BAD=∠CAE=90°BF=EF求证:AF⊥CD
已知:AB=AD,AC=AE,∠BAD=∠CAE=90°BF=EF求证:AF⊥CD

已知:AB=AD,AC=AE,∠BAD=∠CAE=90°BF=EF求证:AF⊥CD
证明：延长AF到P,使AF=FP
∠AFB=∠PFE（对顶角）
∵BF=EF
∴△ABF≌△PEF
∴AB=PE,∠ABE=∠PEF
∴∠AEP=∠AEB+∠∠PEF
又∵∠BAD=∠CAE=90°,∴∠BAC=∠EAD
∴∠ABE +∠BAC+∠AEB=∠EAD+∠CAD
∴∠CAD=∠ABE +∠AEB=∠AEP
又∵AB=AD,AC=AE
∴在△ACD和△APE中
AC=AP,∠CAD=∠AEP,AB=PE
∴△ ACD≌△APE
∴∠C=∠PAD
∠PAD+∠CAF=90°
∠C+∠CAF=90°,
∴AF⊥CD