tag:blogger.com,1999:blog-6933544261975483399.post1268372260308425589..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1394: Three Circles and Three Pair of Common Internal Tangents, ConcurrencyAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-49800342959667466082018-10-16T19:31:31.924-07:002018-10-16T19:31:31.924-07:00https://photos.app.goo.gl/bEcZvA4JTfREGsin9
Let r...https://photos.app.goo.gl/bEcZvA4JTfREGsin9<br /><br />Let rA, rB, rC are radius of circles A, B and C<br />Note that B,F,C are colinear<br />Triangle BB’F similar to CC’F ( case AA)<br />So BF/FC= BB’/CC’= rB/rC<br />Similarly we also have CE/AE= rC/rA and DA/DB= rA/rB<br />Multiply these expressions side by side and simplify it we get<br />BF/BC x EC/EA x DA/DB= 1<br />So AF, BE, CD are concurrent per Ceva’s theorem<br /><br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3125693059664338532018-10-12T05:46:17.689-07:002018-10-12T05:46:17.689-07:00Fie triunghiul format de ABC. Se arată ușor că pun...Fie triunghiul format de ABC. Se arată ușor că punctele E , F și D se găsesc pe laturile AC, CB și AB.<br />Notând cu M punctul de tangență cu cercul A și cu N punctul de tangență comună cu cercul C, din asemănarea triunghiurilor AME și CNE ( AA) se obține raportul AE/EC = r_A/r_C si analog CF/FB = r_c/r_B respectiv BD/DA = r_B/r_A . De aici AE/EC ∙ CF/FB ∙ BD/DA= r_A/r_C ∙ r_A/r_C ∙ r_B/r_A = 1 și conform reciproca teoremei Ceva rezultă că dreptele AF, CD, BE sunt concurente.<br />Florin Popa , Comănești, România.<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87155311320164378962018-10-10T13:05:07.245-07:002018-10-10T13:05:07.245-07:00By ceva's theorem: (AE)/(EC).(CF/CB).(BD/DA)=1...By ceva's theorem: (AE)/(EC).(CF/CB).(BD/DA)=1 (Tr AET similar to tr ECP T,P tg points)<br />(Ra/Rc).(Rc/Rb).(Rb/Ra)=1<br />c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-34297945053088114422018-10-09T18:01:27.831-07:002018-10-09T18:01:27.831-07:00circle A, radius is Ra;
connect AB, ab go through ...circle A, radius is Ra;<br />connect AB, ab go through D,<br />same for AC,BC; E on AC, F on BC;<br />so AE/EC= Ra/Rc, CF/FB=Rc/Rb; BD/DA=Rb/Ra;<br />so AE/EC * CF/FB * BD/DA =1,<br />so AF,BE,CD concurrentAnonymousnoreply@blogger.com