tag:blogger.com,1999:blog-6933544261975483399.post306553409768204669..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 130Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-70165853161242708732020-05-24T13:32:14.748-07:002020-05-24T13:32:14.748-07:00PA/AD=PBA/ABD=PCA/ACD=(PBA+PCA)/(ABD+ACD)=(PBA+PCA...PA/AD=PBA/ABD=PCA/ACD=(PBA+PCA)/(ABD+ACD)=(PBA+PCA)/ABC<br />PB/BE=PBC/BEC=PBA/BEA=(PBC+PBA)/(BEC+BEA)=(PBC+PBA)/ABC<br />PC/CF=PBC/CFB=PAC/CAF=(PBC+PAC)/(CFB+CAF)=(PBC+PAC)/ABC<br />Therefore<br />PA/AD+PB/BE+PC/CF=(PBA+PCA)/ABC+(PBC+PBA)/ABC+(PBC+PAC)/ABC<br />=(2PBA+2PCA+2PBC)/ABC=2(PBA+PCA+PBC)/ABC=2ABC/ABC=2rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39648563053570327082010-02-10T10:13:25.041-08:002010-02-10T10:13:25.041-08:00from an cevian theorem
PD/AD + PE/BE + PF/FC = 1
...from an cevian theorem<br /><br />PD/AD + PE/BE + PF/FC = 1<br />=><br />(AD - PA)/AD + (BE - PB)/BE + (FC - PC)/FC = 1<br /><br />AD/AD - PA/AD + BE/BE - PB/BE + FC/FC - PC/FC = 1<br /><br />1 - PA/AD + 1 - PB/BE + 1 - PC/FC = 1<br /><br />PA/AD + PB/BE + PC/FC = 2<br />-------------------------------------------c .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3199900707664927662009-12-16T09:38:45.477-08:002009-12-16T09:38:45.477-08:00A Cevian is any line segment which joins a vertex ...A Cevian is any line segment which joins a vertex of a triangle with a point on the opposite side (or its extension).Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52874189874214369542009-12-14T12:09:58.601-08:002009-12-14T12:09:58.601-08:00can I know what is a cevian, pleasecan I know what is a cevian, pleasec .t . e. onoreply@blogger.com