Question

如图,在三角形ABC中,∠ABC的平分线BF与三角形ABC的外角∠ACE的平分线CD相交于点D,若∠A=40度

Answer 1

∵∠ABC的平分线BF与△ACB的外角∠ACE的平分线CD相交于点D,

∴∠DCE=1/2 ∠ACE,∠DBC=1/2 ∠ACE,

∵∠DCE是△BCD的外角,

∴∠D=∠DCE-∠DBC

=1/2 ∠ACE- 1/2∠ABC

=1/2 （∠A+∠ABC）-1/2 ∠ABC

=1/2 ∠A+ 1/2∠ABC- 1/2∠ABC

= 1/2∠A= ×40°

=20°．

(2)

∠A=2∠D