如图,BP平分∠BFC,CP平分∠ECB

∵∠ACD=∠A+∠ABC,CP平分∠ACD∴∠PCD=∠ACD/2=（∠A+∠ABC）/2∵BP平分∠ABC∴∠PBC=∠ABC/2∴∠PCD=∠P+∠PBC=∠P+∠ABC/2∴∠P+∠ABC/2

根据三角形外角的性质,有∠ACD=∠A+∠ABC,∠PCD=∠P+∠PBC而,BP、CP分别是∠ABC、∠ACD的平分线,即有,∠PBC=(1/2)*∠ABC,∠PCD=(1/2)*∠ACD代入化简得

1）∵BP平分∠CBD,∴点P到BC、BD的距离相等（角平分线上的点到这个角两边的距离相等）同理,∵CP平分∠BCE,∴点P到CB、CE的距离相等,∴点P到BD和CE（即AB、AC）的距离相等,∴点P

∠BPC=115度

证明：过点P分别过点P作PD⊥AM于D,PE⊥BC于E,PF⊥AN于F．∵BP、CP是△ABC的外角平分线,∴PD=PE,PE=PF,∴PD=PF．∴点P必在∠BAC的平分线上．（到角两边距离相等的点

从P点分别作BC、AC、AB直线上的垂线,然后就可以证明三条线相等（平分线）了,然后直接得到P在∠BAC的平分线上.

∠BPC+∠PBC+∠PCB=180∠BPC+1/2∠ABC+1/2∠ACB=180（1）∠A+∠ABC+∠ACB=1801/2∠A+1/2∠ABC+1/2∠ACB=90（2）（1）—(2)得:∠BP

设∠ABP=∠CBP=∠1,∠ACP=∠BCP=∠2,由△ABC：∠A=180°-2∠1-2∠2（1）由△PBC：∠BPC=∠P=180-∠1-∠2（2）（2）×2-（1）得：2∠P-∠A=180°∴

如图,bp、cp分别平分∠abc和∠acd,且bp与cp相交于点p,∠p与∠a有着什么样的数量关系

∵∠ACD＝∠A+∠ABC,CP平分∠ACD∴∠PCD＝∠ACD/2＝（∠A+∠ABC）/2∵BP平分∠ABC∴∠PBC＝∠ABC/2∴∠PCD＝∠P+∠PBC＝∠P+∠ABC/2∴∠P+∠ABC/2

/>∵∠ACD＝∠A+∠ABC,CP平分∠ACD∴∠PCD＝∠ACD/2＝（∠A+∠ABC）/2∵BP平分∠ABC∴∠PBC＝∠ABC/2∴∠PCD＝∠P+∠PBC＝∠P+∠ABC/2∴∠P+∠ABC

∵∠ACD＝∠A+∠ABC,CP平分∠ACD∴∠PCD＝∠ACD/2＝（∠A+∠ABC）/2∵BP平分∠ABC∴∠PBC＝∠ABC/2∴∠PCD＝∠P+∠PBC＝∠P+∠ABC/2∴∠P+∠ABC/2

/>∵∠ACD＝∠A+∠ABC,CP平分∠ACD∴∠PCD＝∠ACD/2＝（∠A+∠ABC）/2∵BP平分∠ABC∴∠PBC＝∠ABC/2∴∠PCD＝∠P+∠PBC＝∠P+∠ABC/2∴∠P+∠ABC

1.角p等于65度,角P等于角A加角D2,解不出来····3.不符合

∠A=50,所以∠ABC+∠ACB=130∠ACP=1/2（180-∠ACB）=90-∠ACB/2∠P=180-∠PBC-（∠ACB+∠ACP）因为∠PBC=∠ABC/2所以∠P=180-∠ABC/2

∵∠BCP=12∠BCE=12（∠A+∠CBA），∠CBP=12∠CBD=12（∠A+∠ACB）；（角平分线的定义及三角形的一个外角等于与它不相邻的两个内角的和）∴∠BCP+∠CBP=∠A+12（∠C

∵∠A=86°,∴∠ABC+∠ACB=94°又∵BP平分∠ABC,CP平分∠ACB∴∠PBC=1/2∠ABC,∠PCB=1/2∠ACB.∴∠PBC+∠PCB=1/1(∠ABC+∠ACB）=47°.∴∠

∵在△ABC中，∠A=50°，∴∠ABC+∠ACB=180°-50°=130°．∵BP平分∠ABC，CP平分∠ACB，∴∠PBC+∠PCB=12（∠ABC+∠ACB）=12×130°=65°，∴∠BP

在△BCP中,∵∠PBC+∠P+∠PCB=180°∴∠P=180°-1/2∠ABC-（∠PCA+∠ACB)=180°-1/2∠ABC-(1/2∠ACD+∠ACB)=180°-1/2∠ABC-[1/2(

根据题意，∠PCD=∠P+∠PBC，∠ACD=∠A+∠ABC，∵BP平分∠ABC，CP平分∠ABC的外角∠ACD，∴∠ABC=2∠PBC，∠ACD=2∠PCD，∴∠A+∠ABC=2（∠P+∠PBC），