Question

1如图,BD、CD分别是△ABC的外角∠EBC与∠FCB的平分线,BD、CD相交于点D求证：∠D=90°-&frac12

Answer 1

（1）设∠ABC=x,∠ACB=y,则∠A=180-x-y∠EBD=∠CBD=1/2(180-x)=90-½X∠FCD=∠BCD=1/2(180-y)=90-½y∠D=180-∠EBD-∠FCD=½X+½y=90°-½∠A（2）延长AD交BC于E 则∠AEC=∠A+∠B∠D=...