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Geometry Difficulty 3.7 AMC 10/12 Prove it North Macedonia

Let ABCDABCD be a parallelogram and let EE, FF, GG and HH be the midpoints of the sides ABAB, BCBC, CDCD and DADA, respectively. If BH∩AC=IBH \cap AC = I, BD∩EC=JBD \cap EC = J, AC∩DF=KAC \cap DF = K and AG∩BD=LAG \cap BD = L, then prove that the quadrilateral IJKLIJKL is a parallelogram.

Solution

Let AC∩BD=OAC \cap BD = O. Clearly, AOAO and BHBH are medians in the triangle ABDABD, hence II is the centroid of ABDABD. Similarly KK is the centroid of BCDBCD. If IO‾=x\overline{IO} = x, then AI‾=2x\overline{AI} = 2x. Similarly, if KO‾=y\overline{KO} = y, then CK‾=2y\overline{CK} = 2y. Therefore 3x=AO‾=CO‾=3y3x = \overline{AO} = \overline{CO} = 3y, i.e. x=yx = y. We analogously prove that JO‾=LO‾\overline{JO} = \overline{LO}. It follows that IJKLIJKL is a parallelogram.

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