如图,连接CP,求PC平分角BPE

连结OP∴∠OCP=∠OPC=∠DCP∴OP//CD∵CD⊥AB∴OP⊥AB∴∴P是弧AB中点

∵∠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

一、角FAC=180度-角BAC角ACE=180度-角ACB角FAC+角ACE=360度-(角BAC+角ACB)角ACB+角BAC=180度-角B=180度-70度=110度角FAC+角ACE=360

∠BPC=115度

如下：∠ACD=∠ABC+∠A=∠ABC+70°∠PCD=1/2*∠ACD=1/2*∠ABC+35°∠PCD=∠PBC+∠P∠PBC+∠P=1/2*∠ABC+35°∠P=35°

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°∴

∵∠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

证明：（1）∵PC是直径,∴∠PDC=90°,∴∠BDP+∠ADC=90°,又∠BDP=∠DCP,∴∠ADC=∠ACD,即AC=AD,∴AD也是⊙O的切线．∴BD2=BP•BC,∵BD=2

因为∠A=∠B=90°所以AD平行BC所以∠ADC+∠BCD=180°因为DP、CP分别平分∠ADC、∠BDC所以∠PDC=二分之一∠ADC,∠PCD=二分之一∠BCD所以∠PDC+∠PCD=90°因

将△ABP绕A点逆时针旋转90°,然后连接PQ,则AQ=AP=1,CQ=PB=3,∠QAC=∠PAB,又∵∠PAB+∠PAC=90°,所以∠PAQ=∠QAC+∠CAP=∠PAB+∠PAC=90°,所以

∠PCD为△PBC外角,故①∠PCD=∠PBC+∠BPC∠ACD为△ABC外角,故②∠ACD=∠ABC+∠BAC将①式乘以2得2∠PCD=2∠PBC+2∠BPC...③其中2∠PCD=∠ACD.④2∠

∠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

根据已知条件可以得出,三角形AFD是等腰三角形,角FDA=角FAD(因为EF垂直平分AD,假设EF与AD的交点为O,则AO=DO,且角AOF=角DOF）,根据三角形原则:角ADF=角B+角DAB,角F

∵∠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°.∴∠

∵BP平分角ABC,CP平分角ACB,∴∠ABP=1/2∠ABC=40°,∠ACP=1/2∠ACB=25°,延长BP交AC于D,则∠BPC=∠PDC+∠ACP=(∠A+∠ABP)+∠ACP=∠A+40

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(

等于的.因为EF垂直平分AD,所以AE=DE,角ABF=角DEF.又因为EF=EF,所以三角形AEF全等于三角形DEF.所以角EAF=角EDF,AD平分角BAC所以角BAD=角CAD.又角EDF=角B