∠BAC=180-∠B-∠ACB
AD平分∠BAC,
所以,∠DAE=∠BAC/2=(180-∠B-∠ACB)/2=90-∠B/2-∠ACB/2
AD⊥EF,∠AEF=90-∠DAE=90-(90-∠B/2-∠ACB/2)=∠B/2+∠ACB/2
∠AEF是三角形BEM外角
所以∠M=∠AEF-∠B=∠B/2+∠ACB/2-∠B=∠ACB/2-∠B/2=(∠ACB-∠B)/2
∠BAC=180-∠B-∠ACB
AD平分∠BAC,
所以,∠DAE=∠BAC/2=(180-∠B-∠ACB)/2=90-∠B/2-∠ACB/2
AD⊥EF,∠AEF=90-∠DAE=90-(90-∠B/2-∠ACB/2)=∠B/2+∠ACB/2
∠AEF是三角形BEM外角
所以∠M=∠AEF-∠B=∠B/2+∠ACB/2-∠B=∠ACB/2-∠B/2=(∠ACB-∠B)/2