∵∠A+∠ABC+∠ACB=180
∴∠ABC+∠ACB=180-∠A
∵∠CBE=180-∠ABC,BD平分∠CBE
∴∠CBD=∠CBE/2=(180-∠ABC)/2=90-∠ABC/2
∵∠BCF=180-∠ACB,CD平分∠BCF
∴∠BCD=∠BCF/2=(180-∠ACB)/2=90-∠ABC/2
∴∠BDC=180-(CBD+∠BCD)
=180-(90-∠ABC/2+90-∠ACB/2)
=∠ABC/2+∠ACB/2
=(∠ABC+∠ACB)/2
=(180-∠A)/2
=90-∠A/2