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.

∵∠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
∠ABC,
∵∠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
×40°
=20°.