急如图2,在三角形abc中,bd和cd分别是∠abc和∠ace的平分线,角a=62°,求∠bdc的度数

1、
∵∠A=62°
∴∠ABC+∠ACB=180°-62°=118°
∵∠DBA=∠DBC=1/2∠ABC,∠DCA=∠DCB=1/2∠ACB
∴∠DBC+∠DCB=1/2∠ABC+1/2∠ACB=1/2(∠ABC+∠ACB)=1/2*118°=59°
∴∠BDC=180°-(∠DBC+∠DCB)=180°-59°=121°
2、
∵∠A=62°
∴∠ABC+∠ACB=180°-62°=118°
∴∠EBC+∠BCF=(180°-∠ABC)+(180°-∠ACB)=360°-(∠ABC+∠ACB)=242°
∵∠DBE=∠DBC=1/2∠EBC,∠DCF=∠DCB=1/2∠FCB
∴∠DBC+∠DCB=1/2∠EBC+1/2∠FCB=1/2(∠EBC+∠FCB)=1/2*242°=121°
∴∠BDC=180°-(∠DBC+∠DCB)=180°-121°=59°
3、
∵∠ACE=∠A+∠ABC,∠ACD=∠DCE=1/2∠ACE
∴∠DCE=1/2(∠A+∠ABC)=1/2∠A+1/2∠ABC
∵∠ABD=∠DBC=1/2∠ABC,∠DCE=∠DBC+∠BDC
∴∠BDC=1/2∠A
∵∠A=62°
∴∠BDC=31°