tag:blogger.com,1999:blog-6933544261975483399.post1571473068208217952..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 829: Triangle, Circumcircle, Parallel, Chord, Side, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-65539662112737198782017-06-06T23:27:41.401-07:002017-06-06T23:27:41.401-07:00Problem 8Since ACED is an isosceles trapezoid, AE ...Problem 8Since ACED is an isosceles trapezoid, AE = CD = y (say).<br /> <br />Let BF = p, DF = q, BG = u and GE = v<br /> <br />From similar triangles,<br /> <br />∆ABF & ∆DCF ; y/5 = q/1 = 5/p…..(1) and<br />∆BCG & ∆AEG; y/x = v/2 = 4/u ….(2)<br /> <br />Now since AC//DE ; p/q = u/v and substituting for p, q, u & v from (1) & (2) we have<br />(25/y)/(y/5) = (4x/y)/(2y/x) or x2 = 125/2 which gives us x = 5√(5/2)<br /> <br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11039930490446404602012-12-01T22:18:34.729-08:002012-12-01T22:18:34.729-08:00Since DE//AC so Arc(AD)=Arc(EC) => angle (ABF)=...Since DE//AC so Arc(AD)=Arc(EC) => angle (ABF)=angle(CBG)<br />So in triangle ABC ,line BF and BG are isogonal to each other.<br />Per Steiner’s theorem for isogonal lines BF and BG <br />(AF/FC).(AG/GC)=AB^2/BC^2<br />So 1/5 . 4/2 =25/x^2<br />So x=5.sqrt(5/2)<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79861031585432657542012-12-01T16:40:25.447-08:002012-12-01T16:40:25.447-08:00Denote BF, BG by p & q resply. Use Stewart'...Denote BF, BG by p & q resply. Use Stewart's Theorem to obtain: 125+x^2=6(p^2+5), 50+4x^2=6(q^2+8)and p/(p+5/p) =q/(q+8/q).<br />Solve simultaneously to get x=5√(2.5) --- same as AtharvaAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77466128562166723832012-12-01T07:09:54.295-08:002012-12-01T07:09:54.295-08:00x=5sqrt(5/2)x=5sqrt(5/2)Anonymoushttps://www.blogger.com/profile/02636256378320598083noreply@blogger.com