tag:blogger.com,1999:blog-6933544261975483399.post2712347932764041724..comments2024-05-19T16:50:59.263-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 128Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-56129987950348626032012-06-13T16:53:17.172-07:002012-06-13T16:53:17.172-07:00To Anonymous: My previous question is already solv...To Anonymous: My previous question is already solved. I've learned more about Van Aubel. Thanks.Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42994095298651038682012-06-13T08:08:20.991-07:002012-06-13T08:08:20.991-07:00There is a simpler proof using areas:
ID/AD
= (ID...There is a simpler proof using areas:<br />ID/AD <br />= (IDB)/(ADB)= (IDC)/(ADC)<br />= [(IDB)+(IDC)]/[(ADB)+(ADC)]=(IBC)/(ABC)<br />= (IBC)/(ABC)<br />Similarly <br />IE/BE = (IAC)/(ABC) and<br />IF/CF = (IAB)/(ABC)<br />Hence <br />ID/AD + IE/BE + IF/CF <br />=[(IBC)+(IAC)+(IAB)]/(ABC)=ABC)/(ABC) <br />= 1<br />Note:In fact PD/AD + PE/BE + PF/CF = 1 <br />holds for any three concurrent cevians APD, BPE, CPFPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-85056006768007665482012-06-12T18:50:24.616-07:002012-06-12T18:50:24.616-07:00To Anonymous:
As far as I know the theorem of Van ...To Anonymous:<br />As far as I know the theorem of Van Aubel, I can't see how it is applyed in your solution posted in April 13, 2012. Could you please explain it?<br />Thanks.Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89645074537304854672012-04-13T05:41:19.566-07:002012-04-13T05:41:19.566-07:00using theorem of van aubel we have :
AI\ID = AF\FB...using theorem of van aubel we have :<br />AI\ID = AF\FB + AE\EC = AC\BC + AB\BC = (AC+AB)\BC<br />now AI\ID + 1 = AD\ID = s\BC (and s = AB + AC + BC)<br />so : ID\AD = BC\s<br />and similar after sum we get : ID\AD + IE\BE + IF\FC = BC\s + AC\s + AB\s = s\s = 1 .<br />so we are done !!<br />adil azrou from morocco .Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78135105043063512852008-07-05T18:09:00.000-07:002008-07-05T18:09:00.000-07:00BI is a bisector of angle BAI/ID = AB/BDCI is a bi...BI is a bisector of angle B<BR/>AI/ID = AB/BD<BR/>CI is a bisector of angle C<BR/>AI/ID = AC/CD<BR/>AI/ID = AB/BD = AC/CD<BR/>AI/ID = ( AB + AC )/( BD + CD )<BR/>AI/ID = ( AB + AC )/BC<BR/>( AI + ID )/ID = ( AB + BC + CA )/BC<BR/>ID/AD = BC/( AB + BC + CA )<BR/>Similarly<BR/>IE/BE = AC/( AB + BC + CA )<BR/>IF/CF = AB/( AB + BC + CA )<BR/>Therefore by addition<BR/>ID/AD + IE/BE + IF/CF = 1<BR/>Magdy EssaftyAnonymousnoreply@blogger.com