俊狼猎英团队为您解答:
∵BD、CD平分∠CBE、∠BCF,
∴∠DBC+∠DCB=1/2∠CBE+1/2∠BCF
=1/2(∠A+∠ACB+∠A+∠ABC)
=1/2(∠A+180°)
=1/2∠A+90°,
∴∠BDC=180°-(∠DBC+∠DCB)
=180°-1/2∠A-90°
=90°-1/2∠A。
∠BDC=180-1/2(∠CBE+∠BCF)
=180°-1/2(2∠A+∠ABC+∠ACB)
=180°-1/2*∠A-1/2(∠A+∠ABC+∠ACB)
=180°-1/2*∠A-1/2*180°
=180°-1/2∠A-90°
=90°-∠A/2