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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1192.
From similar triangles, BF/HD = FE / EH = CF/AH and since BF = CF it follows that AH = HD.Also from similar triangles BF/AH = GF/GH = FC/HD since F and H are respectively the midpoints of BC and AD. 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.Thus we have proved that in trapezoid ABCD, GE bisects the parallel sides.This is also true for trapezoids ABFH and CDFH.Hence AK=KH=HM=MDHence S(KGM) = 2 X S(AGK) = 20Sumith PeirisMoratuwaSri Lanka