如图,在三角形ABC中∠ABC的平分线和三角形ABC的外角平分线交于点D,∠A=90°,求∠D=?

Question

如图,在三角形ABC中∠ABC的平分线和三角形ABC的外角平分线交于点D,∠A=90°,求∠D=?
 

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=1/2 ×90°
=45°．