如图,PA,PB分别切圆O与AB两点P C满足AB·PB-AC· PC=AB·PC-AC·PB且AP⊥PC∠PAB=2∠BPC求∠ACB ,
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![如图,PA,PB分别切圆O与AB两点P C满足AB·PB-AC· PC=AB·PC-AC·PB且AP⊥PC∠PAB=2∠BPC求∠ACB ,](/uploads/image/z/2119978-10-8.jpg?t=%E5%A6%82%E5%9B%BE%2CPA%2CPB%E5%88%86%E5%88%AB%E5%88%87%E5%9C%86O%E4%B8%8EAB%E4%B8%A4%E7%82%B9P+C%E6%BB%A1%E8%B6%B3AB%C2%B7PB-AC%C2%B7+PC%3DAB%C2%B7PC-AC%C2%B7PB%E4%B8%94AP%E2%8A%A5PC%E2%88%A0PAB%3D2%E2%88%A0BPC%E6%B1%82%E2%88%A0ACB+%2C)
如图,PA,PB分别切圆O与AB两点P C满足AB·PB-AC· PC=AB·PC-AC·PB且AP⊥PC∠PAB=2∠BPC求∠ACB ,
如图,PA,PB分别切圆O与AB两点
P C满足AB·PB-AC· PC=AB·PC-AC·PB
且AP⊥PC
∠PAB=2∠BPC
求∠ACB ,
如图,PA,PB分别切圆O与AB两点P C满足AB·PB-AC· PC=AB·PC-AC·PB且AP⊥PC∠PAB=2∠BPC求∠ACB ,
证法1:
AB·PB-AC· PC=AB·PC-AC·PB
(AB+AC)PB=(AB+AC)PC
PB=PC;
∵PA,PB为切线
∴PA=PB=PC;
∵AP⊥PC
∴∠PAC=∠PCA=45°
∠PAB=∠PBA
∠APB=180-2∠PAB;
∠BPC=90-∠APB=90-(180-2∠PAB)=2∠PAB-90°
∵∠PAB=2∠BPC
1/2∠PAB=2∠PAB-90°
∠PAB=60°
∠BPC=1/2∠PAB=30°
∠PCB=∠PBC=1/2(180-∠BPC)=75°
∴∠ACB=∠PCB-∠PCA=75-45=30°;
证法2:
AB·PB-AC· PC=AB·PC-AC·PB
(AB+AC)PB=(AB+AC)PC
PB=PC;
∵PA,PB为切线
∴PA=PB=PC;
∴ABC在P点为圆心PA为半径的圆上;
∴∠ACB=1/2∠APB(同弧所对的圆周角是圆心角的一半)
∠PAB=∠PBA
∠APB=180-2∠PAB;
∵AP⊥PC
∠BPC=90-∠APB=90-(180-2∠PAB)=2∠PAB-90°
∵∠PAB=2∠BPC
1/2∠PAB=2∠PAB-90°
∠PAB=60°
∴∠ACB=1/2*60=30°