BD、CD分别是三角形ABC的两个外角∠CBE、∠BCF的角平分线.求证：∠BDC＝90°-½∠A

∵DB平分∠EBC
∴∠DBC=½∠EBC
∵DC平分∠FCB
∴∠DCB=½∠FCB
∴∠DBC+∠DCB=½∠EBC+½∠FCB
∵∠EBC=∠A+∠ACB
∠FCB=∠A+∠ABC
∴∠EBC+∠FCB=∠A+∠BAC+∠ABC+∠A=180°+∠A
∴∠DBC+∠DCB=90°+½∠A
∴∠BDC=90°-½∠A