AD平分角OAE证明:延长AO交圆于F,连CF有∠AFC=∠ABCAF是直径,所以∠ACF是直角,∠FAC=90-∠AFCAE⊥BC ---> ∠BAE=90-∠ABC=90-∠AFC=∠FAC又因为D为BC弧的中点所以,∠BAD=∠CAD∠BAD-∠BAE=∠CAD-∠FAC∠DAE=∠DAF所以...