∠P=40°,BP.CP分别平分∠ABC和∠ACD,求角A的度数
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/07 12:06:06
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
以PA为边长作等边△PAD,连结BD∵∠PAD=60°=∠BAC∴∠BAD=∠PAC∵AD=AP,AB=AC∴△ABD≌△APC∴BD=PC=5∵PD=PA=3,PB=4∴∠BPD=90°∵∠APD=
∵∠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
过C 做 ∠ACB的角分线 把下面红线带入上面的红线
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/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
这个算一下就好了啊.∠PBC=1/2(∠A+∠ACB)∠PCB=1/2(∠A+∠ABC)∠P=180°-上面两个也就是∠P=180°-∠A-1/2∠ACB-1/2∠ABC因为1/2∠ACB+1/2∠A
∠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
初一,正好学到△的一个外角等于不相邻的两个内角和这道题就反复用这个知识点∵∠P+∠PBC=∠PCD∠A+∠ABC=∠ACD又∠ACD=2∠PCD∴∠A+∠ABC=2(∠P+∠PBC)∴∠A=2∠P+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
证明:作PM⊥AB,交AB延长线于M,PN⊥AC,交AC延长线于N,作PO⊥BC于O∵PB是∠MBC的平分线∴PM=PO【角分线上的点到两边的距离相等】∵PC是∠NCB的平分线∴PN=PO∴PM=PN
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
∠BPC=90-∠A/2∵∠DBC=180-∠ABC,BP平分∠CBD∴∠PBC=∠CBD/2=(180-∠ABC)/2=90-∠ABC/2∵∠BCE=180-∠ACB,CP平分∠BCE∴∠PCB=∠
∠E= ∠AFC=90°(1)∠BAE+∠EAC=90°∠BAE+∠ABE=90°所以∠ABE=∠EAC同理∠BAE=∠ACF(2)AB=AC △ABE≌△CAF(AAS)AE-A