tag:blogger.com,1999:blog-6933544261975483399.post7548341406907567952..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 488: Triangle, Cevian, Concurrency, Circle, CircumcircleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-33346891149110387942013-04-10T15:13:58.818-07:002013-04-10T15:13:58.818-07:00The concurrency point is called "the cyclocev...The concurrency point is called "the cyclocevian conjugate of P".Anonymoushttps://www.blogger.com/profile/05208522114695973019noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-14654827557189516492010-07-31T14:00:45.879-07:002010-07-31T14:00:45.879-07:00Let n=BA’ .BA” = BC’.BC” ( circumcircle A’B’C’ ...Let n=BA’ .BA” = BC’.BC” ( circumcircle A’B’C’ and secants BC’C” , BA”A’ )<br /> and m=AB’.AB”=AC’.AC” ; p=CA’.CA”=CB”.CB’ ( same reasons as above)<br />1. Apply Ceva’s theorem for triangle ABC and point of concurrent D<br />(A’B/A’C) * (B’C/B’A) *(C’A/C’B)=1<br /><br />2. Replace A’B=m/A”B ; A’C=p/A”C etc…. in above equation we have :<br />(m/A”B)/(p/A”C) * (p/B”C)/(m/B”A) * ( m/C”A)/(n/C”B) =1 <br />3. Simplify above equation we get (A”C/A”B )*(B”A/B”C)*(C”B/C”A) =1 <br />Per Ceva’s theorem AA”, BB”, CC” are concurrent.<br /><br />Peter Tran<br />vstran@yahoo.comPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com