Question

在三角形ABC中,角ABC与角ACB的外角平分线交于点D,且角A=50度,求角D的度数

Answer 1

∠DCA=1/2∠ECA=1/2(∠A+∠ABC)

∠D=180-(∠DCA+∠ACB+∠DBC)

=180-[1/2(∠A+∠ABC)+1/2∠ABC+∠ACB]

=180-(1/2∠A+∠ABC+∠ACB)=

=1/2∠A

=25°