∵∠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°．