做圆上到该圆上两点距离为任意比值n(如3.50)的点

ataorj

做圆上到该圆上两点距离为任意比值n(如3.50)的点
即如图:已知A,B;求做C,D
做法如图 知两点求1点-6.gsp:AE=AB/(n+1),AP=AE/(n-1),PE为半径可得C,D
请证明做法成立,并证明PC,PD都是圆O切线

ccmmjj说"其实这只是阿波罗圆的应用，对任何比值都适用。"见:
【几何作图】知两点，求1点，作法多种……
http://www.mathchina.com/bbs/for ... 82&fromuid=4464
上面没证明.

ataorj

证明:
(PA+AE)=AE[1/(n-1)+1]=AB/(n+1)[1/(n-1)+1]=nAB/(n+1)/(n-1)
(PA+AB)=AB/(n+1)/(n-1)+AB=AB[1/(n+1)/(n-1)+1]=nnAB/(n+1)/(n-1)
PE/PB=(PA+AE)/(PA+AB)=1/n
BC上取F,使得EF//PC,则EF/PC=BF/BC=BE/BP=(BP-PE)/BP=1-PE/PB=1-1/n,
则F为CB的n等分点.
由EF/PC=1-1/n,得EF=PC(n-1)/n=PE(n-1)/n=(PA+AE)(n-1)/n=nAB/(n+1)/(n-1)*(n-1)/n=AB/(n+1)
而AE=AB/(n+1),则EF=AE,则△CEA≌△CEF,CA=CF.则CB/CA=CB/CF=n
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PC是圆O切线是要证PC^2=PA*PB,证如下:
PC=PE=(PA+AE)=nAB/(n+1)/(n-1)
PA*PB=PA*(PA+AB)=[AB/(n+1)/(n-1)][nnAB/(n+1)/(n-1)]=[nAB/(n+1)/(n-1)]^2=PC^2
PD是圆O切线的证明略.