tag:blogger.com,1999:blog-6933544261975483399.post4519988574872780959..comments2022-12-02T06:23:48.816-08:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1192: Trapezoid, Triangle, Diagonal, Midpoint, Area, ParallelAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-5031661290140566392016-02-28T16:22:30.328-08:002016-02-28T16:22:30.328-08:00From similar triangles, BF/HD = FE / EH = CF/AH an...From similar triangles, BF/HD = FE / EH = CF/AH and since BF = CF it follows that AH = HD.<br /><br />Also from similar triangles BF/AH = GF/GH = FC/HD since F and H are respectively the midpoints of BC and AD. <br /><br />This (and since < GFC = < HDG) implies that triangles GFC and GHD are similar and so < FCG = < HDG thus implying that D,C,G are collinear points.<br /><br />Thus we have proved that in trapezoid ABCD, GE bisects the parallel sides.<br /><br />This is also true for trapezoids ABFH and CDFH.<br /><br />Hence AK=KH=HM=MD<br /><br />Hence S(KGM) = 2 X S(AGK) = 20<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com