Question

如图在三角形ABC中,D是角ACB与角ABC的角平分线的交点,BD的延长线交AC于E,且角EDC等于50度,求角A

Answer 1

http://hi.baidu.com/yzwjmx%5F/album/item/4405731e5dafcbd4e1fe0bf1.html

∵∠EDC=∠1+∠2

∵D是∠ACB与∠ABC的角平分线的交点

∴∠EDC=1/2（∠ABC+∠ACB)=1/2(180°-∠A)=90°-1/2∠A

∴50°=90°-1/2∠A

∴∠A=80°