如图，△ABC的两条外角平分线交于点P，求证：∠P=90°-1/2∠A

显示全部楼层 · 2013-9-8 13:22:06

证明：∵∠DBC＝180-∠ABC，BP平分∠DBC∴∠PBC＝∠DBC/2＝（180-∠ABC）/2＝90-∠ABC/2∵∠ECB＝180-∠ACB，CP平分∠FCB∴∠PCB＝∠ECB/2＝（180-∠ACB）/2＝90-∠ACB/2∴∠P＝180-（∠PBC+∠PCB）＝180-（90-∠ABC/2+90-∠ACB/2）＝（∠ABC+∠ACB）/2＝（180-∠A）/2＝90-∠A/2...

千问 · 2013-9-8 13:22:06

∠PBC=1/2(∠A+∠ACB)∠PCB=1/2(∠A+∠ABC)∠P=180°-上面两个也就是 ∠P=180°-∠A-1/2∠ACB-1/2∠ABC因为1/2∠ACB+1/2∠ABC+1/2∠A=90°所以整理一下:∠P=90°—1/2∠A。...