tag:blogger.com,1999:blog-6933544261975483399.post8058679839526155218..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 582: Quadrilateral, Diagonals, Incenters, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-90003665277687993802011-03-06T04:35:58.447-08:002011-03-06T04:35:58.447-08:00angle EOF is right angled as the bisectors of a li...angle EOF is right angled as the bisectors of a linear pair enclose a right angle. so by applying Pythagoras theorem and associativity property of addition, we have EF^2 +GH^2 = (FO^2 + EO^2) + (HO^2 + GO^2) = (FO^2 + GO^2) + (EO^2 + HO^2) = FG^2 + EH^2.<br />Q. E. D.sourav mishranoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-28270966095757392922011-02-23T21:20:42.513-08:002011-02-23T21:20:42.513-08:00Connect OE,OF,OG,OH
We have E,O,G collinear and F,...Connect OE,OF,OG,OH<br />We have E,O,G collinear and F,O,H collinear ( angle bisectors of vertical angles)<br />We also have EG perpendicular to FH at O ( angle bisectors of 2 supplement angles)<br />apply Pythagoras therorem we have<br />EF^2+HG^2=OE^2+OF^2+OH^2+OG^2=EH^2+FG^2<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com