tag:blogger.com,1999:blog-6933544261975483399.post2285940150533366048..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 596: Quadrilateral, Right Triangle, Isosceles, MidpointAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-41873723104886498502019-01-29T00:29:26.959-08:002019-01-29T00:29:26.959-08:00BCDE is a kite, hence CE is the perpendicular bise...BCDE is a kite, hence CE is the perpendicular bisector of BD cutting CE at G say<br /><br />Consider Tr.s EFG and ABC<br /><br />< ABC = < EGF, AB/EG = BC/FG = 2<br /><br />Hence the 2 Tr.s are similar and x =2/2 = 10<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41057388128458943952012-07-19T16:15:40.799-07:002012-07-19T16:15:40.799-07:00http://img849.imageshack.us/img849/1084/problem596...http://img849.imageshack.us/img849/1084/problem596.png<br /><br />Draw lines per attached sketch<br />We have ED=EB and CB=CD => CE is perpendicular bisector of BD<br />∆ EFC ≅ to ∆ EGC …..( Case SAS)<br />Since E and G are midpoints of AD and CD => EG=1/2.AC=10<br />So x=EF=EG=10Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68566235727278089232012-05-26T03:24:22.272-07:002012-05-26T03:24:22.272-07:00By applying Apollonius' theorem to triangle AC...By applying Apollonius' theorem to triangle ACD and BCF to get two equations.<br />Eliminating the length of CD, CF and BF will get 2x = 20.Anonymousnoreply@blogger.com