解:过点E作EG⊥族轿BC交BC与点G∵兆氏肆AD⊥BC于D则有EG∥AD则∠GEB=∠EBA∵BE平分∠CBA交AC于E即∠AEB=∠GEB则核或∠AEB=∠AFE则AE=AF.得证
∠AFE=∠BFD=90-∠DBF∠AEF=∠AEB=90-∠ABE因喊绝丛为∠郑樱DBF=∠ABE所以∠宏弊AFE=∠AEF