延长BE与AD交于F,
∠ACB=90°∠CBA=45°,AC=BC,又CD=CE,∠BCE=∠ACD=RT∠,△ACD≌△BCE
∠CBE=∠CAD,∠CEB=∠FEA,又∠CBE+∠CEB=90°,所以∠CAD+∠FEA=90°,
∠BFA=180°-∠CAD+∠FEA=180°-90°=90°,BF⊥AD.