证明:∠BAP=∠BAC/2=(180-∠B-∠ACB)/2=90-∠B/2-∠ACB/2
∠APC为三角形ABP外角,
所以∠APC=∠B+∠BAP=∠B+90-∠B/2-∠ACB/2=90+∠B/2-∠ACB/2
因为MN⊥AD,所以∠EMB=90-∠APC
即∠EMB=90-(90+∠B/2-∠ACB/2)=∠ACB/2-∠B/2=1/2(∠ACB-∠B)
证明:∠BAP=∠BAC/2=(180-∠B-∠ACB)/2=90-∠B/2-∠ACB/2
∠APC为三角形ABP外角,
所以∠APC=∠B+∠BAP=∠B+90-∠B/2-∠ACB/2=90+∠B/2-∠ACB/2
因为MN⊥AD,所以∠EMB=90-∠APC
即∠EMB=90-(90+∠B/2-∠ACB/2)=∠ACB/2-∠B/2=1/2(∠ACB-∠B)