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Geometry Difficulty 4.8 AIME Prove it North Macedonia

ABCDEABCDE is pentagon where KK, LL, MM, NN are the midpoints of ABAB, BCBC, CDCD, DEDE respectively. Let PP, QQ, FF be the midpoints of KMKM, LNLN, ADAD, respectively. Prove that PQPQ and AEAE are parallel and AE=4PQ\overline{AE} = 4\overline{PQ}.

Solution

The quadrilateral KFMLKFML is parallelogram. The point PP is the midpoint for KMKM and PLFP \in LF and also the midpoint for LFLF. In the triangle LFNLFN, we
Figure 1

have PQFNPQ \parallel FN and PQ=12FN\overline{PQ} = \frac{1}{2}\overline{FN}. In the triangle ADEADE, we have FNAEFN \parallel AE and FN=12AE\overline{FN} = \frac{1}{2}\overline{AE}. Finally PQFNAEPQ \parallel FN \parallel AE and PQ=12FN=14AE\overline{PQ} = \frac{1}{2}\overline{FN} = \frac{1}{4}\overline{AE}, i.e. PQAEPQ \parallel AE and AE=4PQ\overline{AE} = 4\overline{PQ}.

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