BP,CP分别平分角ABD,角ACD若角A=40度,角P 等于多少度
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/21 19:04:02
因为,∠BCE=∠A+∠ABC,∠CBD=∠A+∠ACB所以,∠2=1/2*(∠A+∠ABC),∠1=1/2*(∠A+∠ACB)所以,∠BPC=180-(∠1+∠2)=180-1/2*(∠A+∠ACB
过P分别作BM、BN、AC的垂线段PE、PF、PG.∵AP是角MAC的角平分线所以PE=PG同理PF=PG所以PE=PF所以BP平分角MBN
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
1.角p等于65度,角P等于角A加角D2,解不出来····3.不符合
证明:过点P作PM⊥AB于M,PN⊥AC于N,PG⊥BC于G∵PM⊥AB,PG⊥BC,BP平分∠CBD∴PM=PG∵PN⊥AC,PG⊥BC,CP平分∠BCE∴PN=PG∴PM=PN∴AP平分∠BAC
∠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
关系:∠BPC=90°+1/2∠A证明:在ABC中,∠ABC和∠ACB的平分线相交于点P所以∠BPC=180°-(∠PBC+∠PCB)=180°-(1/2∠ABC+1/2∠ACB)=180°-1/2(
过P作PF⊥AC,交AC于F过P作PE⊥BC,交BC延长线于E过P作PG⊥AB,交AB延长线于G因为AP平分∠GAC,所以PG=PF(角平分线上的点到角两边距离相等)因为CP平分∠ACE所以PF=PE
在BC延长线上取点E∵∠A+∠ABC+∠ACB=180∴∠ABC+∠ACB=180-∠A∵∠ACE=180-∠ACB,CP平分∠ACE∴∠PCE=∠ACE/2=(180-∠ACB)/2=90-∠ACB
∠ACM=∠A+ABC∠PCM=∠P+∠PBC已知∠ABC=2∠PBC∠ACM=2∠PCM则2∠PCM=∠A+ABC=∠A+2∠PBC=∠A+2∠PCM-2∠P可求∠A=∠P再问:∠A=∠P?
已知,点P在△ABC的外角平分线BP上,可得:点P到直线AB和直线BC的距离相等;已知,点P在△ABC的外角平分线CP上,可得:点P到直线AC和直线BC的距离相等;所以,点P到直线AB和直线AC的距离
证明:需要做辅助线,三条垂线,第一,过P向AC作垂线垂足为D,过P向AB坐垂线垂足为E,过P向BC做垂线垂足为F.之后根据外角平分线,角ECP和角BCP相等,加上直角和公共边,便可说明三角形ECP和F
如果我没画错的话由题意得∠MBP=∠CBP,∠BCP=∠NCP,∠BAP=∠CAP=a/2∴∠BPC=360°-∠ABP-∠BAC-∠ACP=360°-(180°-∠PBM)-a-(180°-∠PCN
过P依次向AB、BC、CD、AD作垂线,垂足依次为E、F、G、H.∵AP平分∠BAD、PH⊥AH、PE⊥AE,∴PH=PE,又AP=AP,∴Rt△PAH≌Rt△PAE,∴AH=AE.······①∵P