Question

如图,D是△ABC的BC边的延长线上一点,∠ABC与∠ACD的平分线交于点P,若∠A=70°,求∠P.

如图,D是△ABC的BC边的延长线上一点,∠ABC与∠ACD的平分线交于点P,若∠A=70°,求∠P.
请用几何语言回答

Answer 1

∠DCA = ∠A + ∠ABC (三角形外角等于不相邻内角和)

∠DCA
= 2(∠DCP) (角平分线)
= 2(∠P + ∠PBC) (三角形外角等于不相邻内角和)
= 2∠P + 2∠PBC
= 2∠P + ∠ABC (角平分线)

∠P = 1/2∠A = 35 (对比上下两条等式可得)

Answer 2

∠A=70，所以∠ABC+∠ACB=110
∠P=180-∠PBC-∠PCB=180-∠PBC-∠PCA-∠ACB
因为BP平分∠ABC，CP平分∠ACD
所以∠PBC=∠ABC/2，∠PCA=∠ACD/2=（180-∠ACB）/2=90-∠ACB/2
则∠P=180-∠ABC/2-（90-∠ACB/2）-∠ACB
=90-（∠ABC+∠ACB）/2
=90-110/2
=35

Answer 3

记角ABP为a，角ACB为b，角ACP为c，角BCP为x，
a+b+c+x=180,70+2a=2c,2a+70+b=180.由以上三个式子可求的x=35。（以上数字均代表角度）