BC中点为D,角A角平分线交BC于点E,垂足为F.
∠BAD=∠DAE=∠EAF=∠FAC=22.5度
∠CAF=∠EAF,AF⊥EC.所以∠ACF=∠AEF=∠B+∠BAE=∠B+2∠CAF
∠ACF=90-∠CAF
所以∠B=∠CAF
又∠ACF=∠BCA
所以∠BAC=∠AFC=90
BC中点为D,角A角平分线交BC于点E,垂足为F.
∠BAD=∠DAE=∠EAF=∠FAC=22.5度
∠CAF=∠EAF,AF⊥EC.所以∠ACF=∠AEF=∠B+∠BAE=∠B+2∠CAF
∠ACF=90-∠CAF
所以∠B=∠CAF
又∠ACF=∠BCA
所以∠BAC=∠AFC=90