如图所示,在△ABC中,∠ABC的平分线BF与△ACB的外角∠ACE的平分线CD相交于点D,若∠A=40°,试求∠D的度数.
问题描述:
如图所示,在△ABC中,∠ABC的平分线BF与△ACB的外角∠ACE的平分线CD相交于点D,若∠A=40°,试求∠D的度数.
答
∵∠ABC的平分线BF与△ACB的外角∠ACE的平分线CD相交于点D,
∴∠DCE=
∠ACE,∠DBC=1 2
∠ABC,1 2
∵∠DCE是△BCD的外角,
∴∠D=∠DCE-∠DBC
=
∠ACE-1 2
∠ABC1 2
=
(∠A+∠ABC)-1 2
∠ABC1 2
=
∠A+1 2
∠ABC-1 2
∠ABC1 2
=
∠A=1 2
×40°1 2
=20°.