tag:blogger.com,1999:blog-6933544261975483399.post8990990864684750074..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 607: Intersecting Circles, Triangles, Metric Relations, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-69514108565974579952016-02-01T03:56:14.145-08:002016-02-01T03:56:14.145-08:00< BEC = < BCE = < AFC
Hence BAEF is cyc...< BEC = < BCE = < AFC <br /><br />Hence BAEF is cyclic<br /><br />So 8 x 12 = x(x+10)<br /><br />Solving the quadratic for positive x<br /><br />x = 6<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-259526589016149542011-05-30T18:53:16.801-07:002011-05-30T18:53:16.801-07:00Isosceles triangles ACF, BCE are similar.
CF/CE = ...Isosceles triangles ACF, BCE are similar.<br />CF/CE = AC/BC<br />CF.CB = CE.CA, and so A,B,F,E are concyclic.<br />GBA, GFE are secants from G to the circle BAEF.<br />GB.GA = GF.GE,<br />x(x+10)= (8).(12), <br />(x + 5)² = 121,<br />x+5=11, x=6Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.com