tag:blogger.com,1999:blog-6933544261975483399.post2332422954368905843..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1131: Triangle Area, Two Cevians, Equal Product of AreasAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-44091224378712576982018-05-04T09:37:31.109-07:002018-05-04T09:37:31.109-07:00From tr EDB and tr EDC => S1/(S2+S5) = BD/DC
fr...From tr EDB and tr EDC => S1/(S2+S5) = BD/DC<br />from tr ADB and tr ADC => (S1+S2+S4)/(S3+S5) = BD/DC<br />or S1/(S2+S5) = (S1+S2+S4)/(S3+S5)<br />but S4 S5 = S2 S3 => the resultc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39235550319044441722015-11-02T07:53:10.398-08:002015-11-02T07:53:10.398-08:00Lets compute the ratios S/S3 and S1/S2 and prove t...Lets compute the ratios S/S3 and S1/S2 and prove they are equal:<br /><br />S3/ACD = AF/AD (triangles with same height)<br />ACD/S = CD/BC => ACD = (CD/BC)*S<br />S3 = (AF/AD) * ACD = (AF/AD) * (CD/BC) * S<br />S / S3 = (AD / AF) * (BC / CD)<br /><br />S2 / ADE = DF / AD<br />ADE / S1 = AE / BE => ADE = (AE / BE) * S1<br />S2 = (DF/AD) * ADE = (FD / AD) * (AE / BE) * S1<br />S1 / S2 = (AD / DF) * (BE / AE)<br /><br />Next use Menelaus' theorem for the triangle ABD and the line CFE:<br /><br />( BE / AE) * ( AF / DF ) * ( CD / BC ) = 1<br />( BE / AE) * ( AF / DF ) = BC / CD (multiply both sides by AD/AF)<br />( BE / AE) * ( AD / DF ) = (BC / CD) * (AD / AF)<br /><br />These are exactly the expressions above for S1 / S2 and S / S3<br />Therefore S1 / S2 = S / S3<br /><br />Hence S.S2 = S1.S3<br /><br /><br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25764253609577658922015-07-09T10:38:38.245-07:002015-07-09T10:38:38.245-07:00Let BF cut ED and AC at H and G
Observe that H and...Let BF cut ED and AC at H and G<br />Observe that H and G are harmonic conjugate points of F and B<br />So GF/GB=HF/HB<br />But S3/S= GF/GB and S2/S3= HF/HB<br />The result will follow.<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com