Question

如图,在△ABC中,∠ABC和∠ACB的外角平分线相交于点D，∠BDC=50°，求∠A

Answer 1

由题目可知∠DBC+∠DCB=130,∠EBC+∠FCB=2(∠DBC+∠DCB)=260,则∠ABC+∠ACB=180+180-（∠EBC+∠FCB）=360-260=100，则∠A=180-（∠ABC+∠ACB）=180-100=80
∠A=80

Answer 2

∠BDC=50°,
∠DBC+∠DCB=130°,
∠EBC+∠FCB=260°,
∠ABC+∠ACB=360°-260°=100°,
∠A=80°

Answer 3

80度