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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 880.
Let S(PQR) be the area of ΔPQR. Form ΔDEF. Since ΔBFD, ΔFAE and ΔDEC are congruent, and H, G, M are their orthocenters. ThusS(HFD)+S(FGE)+S(DEM) = S(DEF)SoArea of hexagon = 2×S(DEF) = 12
Connect EF, FD and DENote that area(BFD)=Area(AFE)=Area(CED)=Area(EFD)=6Triangles BFD, DEC and FAE are congruence (case SSS)Triangle EDM congruence to triangle FBH ( case ASA)Triangle FGE congruence to triangle BHD ( case ASA)So Area(EGFHDM)= Area(EFD)+Area(FBD)= 12