问题描述：
在三角形ABC的三条角平分线分别为AD,BE,CF,则角ACF+角CBE+角DAC?
问题解答：
我来补答
∵AD平分∠BAC ∴∠DAC=1/2∠BAC
∵BE平分∠ABC ∴∠CBE=1/2∠ABC
∵CF平分∠ACB ∴∠ACF=1/2∠ACB
∵∠BAC+∠ABC+∠ACB=180°
∴∠ACF+∠CBE+∠DAC
=1/2∠BAC+1/2∠ABC+1/2∠ACB
=1/2（∠BAC+∠ABC+∠ACB）
=(1/2)×180°
=90°
∵BE平分∠ABC ∴∠CBE=1/2∠ABC
∵CF平分∠ACB ∴∠ACF=1/2∠ACB
∵∠BAC+∠ABC+∠ACB=180°
∴∠ACF+∠CBE+∠DAC
=1/2∠BAC+1/2∠ABC+1/2∠ACB
=1/2（∠BAC+∠ABC+∠ACB）
=(1/2)×180°
=90°
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