(1):∵BD、CD是∠ABC和∠ACB的角平分线,
∴∠DBC=
1
2 ∠ABC,∠DCB=
1
2 ∠ACB,
∵∠ABC+∠ACB=180°-∠A,
∠BDC=180°-∠DBC-∠DCB=180°-
1
2 (∠ABC+∠ACB)=180°-
1
2 (180°-∠A)=90°+
1
2 ∠A,
∴∠BDC=90°+
1
2 ∠A,
即∠D=90°+
1
2 ∠A.
(2):∵BD、CD分别是∠CBE、∠BCF的平分线
∴∠DBC=
1
2 ∠EBC,∠BCD=
1
2 ∠BCF,
∵∠CBE、∠BCF是△ABC的两个外角
∴∠CBE+∠BCF=360°-(180°-∠A)=180°+∠A
∴∠DBC+∠BCD=
1
2 (∠EBC+∠BCD)=
1
2 (180°+∠A)=90°+
1
2 ∠A,
在△DBC中∠BDC=180°-(∠DBC+∠BCD)=180°-(90°+
1
2 ∠A)=90°-
1
2 ∠A,即∠D=90°-
1
2 ∠A.
(3)∵BD、CD分别为∠ABC、∠ECA的角平分线,
∴∠1=∠DBC=
1
2 ∠ABC,∠2=∠DCE=
1
2 (∠A+∠ABC),
∵∠ACE是△ABC的外角,
∴∠ACE=∠A+∠ABC,
∵∠DCE是△BCD的外角,
∴∠D=∠DCE-∠DBC
=∠DCE-∠1
=
1
2 ∠ACE-
1
2 ∠ABC
=
1
2 (∠A+∠ABC)-
1
2 ∠ABC
=
1
2 ∠A.
故答案为:∠D=90°+
1
2 ∠A;∠D=90°-
1
2 ∠A;∠D=
1
2 ∠A.