如图8,已知:AE⊥AB,AF⊥AC,AE=AB,AF=AC.试证明:(1)CE=BF;(2)CE⊥BF

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如图8,已知:AE⊥AB,AF⊥AC,AE=AB,AF=AC.试证明:(1)CE=BF;(2)CE⊥BF

如图8,已知:AE⊥AB,AF⊥AC,AE=AB,AF=AC.试证明:(1)CE=BF;(2)CE⊥BF
如图8,已知:AE⊥AB,AF⊥AC,AE=AB,AF=AC.试证明:(1)CE=BF;(2)CE⊥BF

如图8,已知:AE⊥AB,AF⊥AC,AE=AB,AF=AC.试证明:(1)CE=BF;(2)CE⊥BF
(1)EAC全等BAF
(2)设EC交AB于M,CE交BF于N,
由EAC全等BAF,得到角AEC=ABF,
又角EMA=BMC,所以角BNM=EAB=90.

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