cp平分角ACE,BP平分角ABC,求角P=二分之一角A
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∵∠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
作∠PCB的平分线交PB于E.∵∠ABE=∠CBE=∠ABC/2、∠ACE=∠BCE=∠ACB/2,∴∠BAE=∠CAE=∠BAC/2.∵∠ACD=∠ACP+∠PCD=2∠PCD、∠ABC=∠ABP+
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
证明:过点A作AF⊥CD于F1、∵AP平分∠BPD,AE⊥BP,AF⊥CD∴AE=AF,PE=PF(角平分线性质),∠AEB=∠AFC∵AB=AC∴△ABE≌△ACF(HL)∴∠ABP=∠ACP2、∵
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°
(1)分别过点P作PD⊥AB于D,PE⊥BC于E,PF⊥AC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.(2)由于角A=50,则角B
∵∠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
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∠
角BPC=180°-角PBC-角PCB=180°-1/2(角ABC+角DCB)=180°-1/2(360°-角A-角D)=180°-180°+1/2(角A+角D)所以:2角BPC=角A+角D
∠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(
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
证明:作PM⊥AB于点M,PN⊥AC于点N,PO⊥BC于点O∵BP平分∠DBC∴PM=PO∵CP平分∠BCE∴PN=PO∴PM=PN∴点在∠A的平分线上
∠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
∵∠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°.∴∠
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