塞瓦定理的应用将△ABC各内角三等分,每两个角的相邻三等分线相交得△PQR,又AX,BY,CZ分别分∠BAC,∠ABC,∠ACB且它们与QR,RP,PQ,交于X,Y,Z.求证:PX,QY,RZ三线共点(最好附图解释)
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![塞瓦定理的应用将△ABC各内角三等分,每两个角的相邻三等分线相交得△PQR,又AX,BY,CZ分别分∠BAC,∠ABC,∠ACB且它们与QR,RP,PQ,交于X,Y,Z.求证:PX,QY,RZ三线共点(最好附图解释)](/uploads/image/z/14297376-48-6.jpg?t=%E5%A1%9E%E7%93%A6%E5%AE%9A%E7%90%86%E7%9A%84%E5%BA%94%E7%94%A8%E5%B0%86%E2%96%B3ABC%E5%90%84%E5%86%85%E8%A7%92%E4%B8%89%E7%AD%89%E5%88%86%2C%E6%AF%8F%E4%B8%A4%E4%B8%AA%E8%A7%92%E7%9A%84%E7%9B%B8%E9%82%BB%E4%B8%89%E7%AD%89%E5%88%86%E7%BA%BF%E7%9B%B8%E4%BA%A4%E5%BE%97%E2%96%B3PQR%2C%E5%8F%88AX%2CBY%2CCZ%E5%88%86%E5%88%AB%E5%88%86%E2%88%A0BAC%2C%E2%88%A0ABC%2C%E2%88%A0ACB%E4%B8%94%E5%AE%83%E4%BB%AC%E4%B8%8EQR%2CRP%2CPQ%2C%E4%BA%A4%E4%BA%8EX%2CY%2CZ.%E6%B1%82%E8%AF%81%EF%BC%9APX%2CQY%2CRZ%E4%B8%89%E7%BA%BF%E5%85%B1%E7%82%B9%28%E6%9C%80%E5%A5%BD%E9%99%84%E5%9B%BE%E8%A7%A3%E9%87%8A%29)
塞瓦定理的应用将△ABC各内角三等分,每两个角的相邻三等分线相交得△PQR,又AX,BY,CZ分别分∠BAC,∠ABC,∠ACB且它们与QR,RP,PQ,交于X,Y,Z.求证:PX,QY,RZ三线共点(最好附图解释)
塞瓦定理的应用
将△ABC各内角三等分,每两个角的相邻三等分线相交得△PQR,又AX,BY,CZ分别分∠BAC,∠ABC,∠ACB且它们与QR,RP,PQ,交于X,Y,Z.求证:PX,QY,RZ三线共点(最好附图解释)
塞瓦定理的应用将△ABC各内角三等分,每两个角的相邻三等分线相交得△PQR,又AX,BY,CZ分别分∠BAC,∠ABC,∠ACB且它们与QR,RP,PQ,交于X,Y,Z.求证:PX,QY,RZ三线共点(最好附图解释)
1) 由AR、AQ三等分∠BAC,AX平分∠BAC可知AX平分∠RAQ,根据角平分线定理可知RX/QX=AR/AQ
2) 同理QZ/PZ=CQ/CP,PY/RY=BP/BR,于是(RX/QX)*(QZ/PZ)*(PY/RY)=(AR/AQ)*(CQ/CP)*(BP/BR)=(AR/BR)*(BP/CP)*(CQ/AQ)
3) 根据正弦定理AR/BR=sin(B/3)/sin(A/3),BP/CP=sin(C/3)/sin(B/3),CQ/AQ=sin(A/3)/sin(C/3),于是(RX/QX)*(QZ/PZ)*(PY/RY)=(AR/BR)*(BP/CP)*(CQ/AQ)=[sin(B/3)/sin(A/3)]*[sin(C/3)/sin(B/3)]*[sin(A/3)/sin(C/3)]=1
4) 根据塞瓦定理的逆定理可知PX、QY、RZ三线共点
关于塞瓦定理请参考:http://baike.baidu.com/view/148207.htm
顺便说一下,△PQR是等边三角形,这是有名的Morley定理