如图.bp评分∠abd,cp平分∠acd
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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
∵∠1=0.5∠DBC=0.5(180°-∠ABC),∠2=0.5∠ECB=0.5(180°-∠ACB)∴∠BPC=180°-(∠1+∠2)=180°-【0.5(180°-∠ABC)+0.5(180°
655540由下面化简得(180-角A)/2=角P(180-(360-(180-角A)/2)=角P)
证明:过点P分别过点P作PD⊥AM于D,PE⊥BC于E,PF⊥AN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.(到角两边距离相等的点
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°∴
过点P作PM⊥AB的延长线,垂足为M,PQ⊥BC,垂足为QPN⊥AC的延长线,垂足为N∵∠MBP=∠QBP,∠PCQ=∠PCN∴PM=PQ,PQ=PN∴PM=PN∴AP平分∠BAC
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
∵∠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
∠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
证明:∵AB⊥BP,AC⊥PC∴∠ABP=∠ACP=90∵AP=AP,BP=CP∴△ABP≌△ACP(HL)∴∠APB=∠APC∵PD=PD∴△BPD≌△CPD(SAS)∴∠BDP=∠CDP
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
∵∠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°.∴∠
∠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=∠