Question

如图所示,在△ABC中,∠A=70°,BD平分∠CBE,CD平分∠BCF,BD与CD相交于点D,求∠D的度数

Answer 1

∠ABC=180°-2∠CBD

∠ABC=180°-∠ACB-∠A

2∠CBD=∠ACB+∠A=∠ACB+70°

∠CBD=∠ACB/2 +5°

∠ACB=180°-2∠BCD

∠ACB=180°-∠ABC-∠A

2∠BCD=∠ABC+∠A=∠ABC+70°

∠BCD=∠ABC/2+35°

∠D=180°-∠CBD-∠BCD

=180°-（∠ACB/2+35°）-（∠ABC/2+35°）

=180°-70°-∠ACB/2-∠ABC/2

=110°-（∠ACB+∠ABC）/2

∠ABC+∠ACB=180°-∠A=110°

∠D =110°-110°/2

=110°-55°

=55°