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