点P是△ABC内角平分线BP与外角平分线CP的交点,请写出∠BPC与∠A的关系

∠PCD=∠PBC+∠BPC=∠PBC+40°；(1)PB平分∠ABC,得∠PBC=∠ABC/2;PC平分∠ACD,得∠PCD=∠ACD/2;代入（1）得∠ACD-∠ABC=80°；在△ABC中,∠B

证明：过P作三边AB、AC、BC的垂线段PD、PE、PF，∵BP是△ABC的外角平分线，PD⊥AD，PF⊥BC，∴PD=PF（角平分线上的点到角两边的距离相等），∵点P在∠BAC的角平分线上，PD⊥A

对的因为∠BAC+∠ABC=2(∠BPC+∠PBC),又因为∠ABC=2∠PBC,所以∠BAC+2∠PBC=2∠BPC+2∠PBC,∠BAC=2∠BPC

过P作PE,PF,PG垂直BA,AC,CD角平分线得PE=PGPF=PG即PE=PFPA=PA所以PEA全等PFAEAP=FAPBPC=PCD-PBC=1/2ACD-1/2ABC=1/2(ACD-AB

∵△ABC的内角平分线BP与外角平分线CP交于P，∴∠PBC=12∠ABC，∠PCD=12∠ACD，∵∠ACD=∠A+∠ABC，∠PCD=∠PBC+∠P，∴12（∠A+∠ABC）=∠PBC+∠P=12

延长BA,做PN⊥BD,PF⊥BA,PM⊥AC,设∠PCD=x°,∵CP平分∠ACD,∴∠ACP=∠PCD=x°,PM=PN,∵BP平分∠ABC,∴∠ABP=∠PBC,PF=PN,∴PF=PM,∵∠B

延长BA,做PN⊥BD,PF⊥BA,PM⊥AC,设∠PCD=x°,∵CP平分∠ACD,∴∠ACP=∠PCD=x°,PM=PN,∵BP平分∠ABC,∴∠ABP=∠PBC,PF=PN,∴PF=PM,∵∠B

∠PCD=∠PBC+∠BPC=∠PBC+40°；(1)PB平分∠ABC,得∠PBC=∠ABC/2;PC平分∠ACD,得∠PCD=∠ACD/2;代入（1）得∠ACD-∠ABC=80°；在△ABC中,∠B

（1）已知BD，CD是内角平分线，∵∠A=30°，∴∠ABC+∠ACB=180°-∠A=180°-30°=150°，∴∠DBC+∠DCB=12（∠ABC+∠ACB）=12×150°=75°，∴∠BDC

∠BPC=1/2∠A列式：∠BPC=1/2C外角-1/2∠ABC=1/2(180-∠ACB-∠ABC)=1/2∠A

延长BA,做PN⊥BD,PF⊥BA,PM⊥AC,设∠PCD=x°,∵CP平分∠ACD,∴∠ACP=∠PCD=x°,PM=PN,∵BP平分∠ABC,∴∠ABP=∠PBC,PF=PN,∴PF=PM,∵∠B

延长BA，作PN⊥BD，PF⊥BA，PM⊥AC，设∠PCD=x°，∵CP平分∠ACD，∴∠ACP=∠PCD=x°，PM=PN，∵BP平分∠ABC，∴∠ABP=∠PBC，PF=PN，∴PF=PM，∵∠B

分两步进行.①先求∠BAC：∠PCD=∠PBC+∠BPC,即1/2∠ACD=40°+1/2∠ABC,∴∠ACD=∠ABC+80°,又∠ACD=∠ABC+∠BAC,∴∠BAC=80°；②证P在∠BAC的

（1）分别过P点别作BC延长线、BE、AC的的垂线,垂足分别为F,H、G因为CP为角ACF的平分线,所以PF=PG因为BP为角EBF的角平分线,所以PF=PH所以PH=PG,AP平分角CAE（2）因为

∠CAB=∠ACD-∠ABC∠PCD=∠PBC+40°∠ACD=2∠PCD=2∠PBC+80°因为∠ABC=2∠PBC,∠ACD=2∠PBC+80°所以∠CAB=2∠PBC+80°-∠ABC=80°

2∠BPC=∠BAC证：∠ACD=∠BAC+ABC=∠BAC+2∠PBC  ∠PCD=∠PBC+∠BPC∵∠acd的平分线cp与内角∠abc的平分线bp交于点p∴∠PCD=∠ACP

115°延长BA,做PN⊥BD,PF⊥BA,PM⊥AC,设∠PCD=x°,∵CP平分∠ACD,∴∠ACP=∠PCD=x°,PM=PN,∵BP平分∠ABC,∴∠ABP=∠PBC,PF=PN,∴PF=PM

延长BA，作PN⊥BD于点N，PF⊥BA于点F，PM⊥AC于点M，设∠PCD=x°，∵CP平分∠ACD，∴∠ACP=∠PCD=x°，PM=PN，∵BP平分∠ABC，∴∠ABP=∠PBC，PF=PN，∴

延伸AQ,交DC于N,∵∠ABD+∠BAD=∠DAP+∠BAD∴∠ABD=∠DAP∴∠PBD=∠DAQ∠MQA=90o∴∠AMP=∠ANB∵∠QDN=∠DBQ+∠BQD,∠AMQ=∠MDQ+∠MQD∴

∵BP、CP是∠ABC和∠ACB的角平分线,∴∠PBC=1/2∠ABC,∠PCB=1/2∠ACB,∵∠ABC+∠ACB=180°-∠A,∠BPC=180°-∠PBC-∠PCB=180°-1/2（∠AB