dear April is it true ? or where I misunderstood?

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Dear planegeometry, I attach here the original problem's statement and its solution, could you please check it for me? Thanks a lot!

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sorry.dear April the right figure is like this:

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LEMMA. – A quadrilateral $ ABCD$ is given and le $ E,\ F$ be, two arbitrary points on $ AC,\ BD$ respectively, such that $ EF\parallel CD$ $ ($ suppose the point $ E,$ between the points $ A,\ Y,$ where $ Y\equiv AC\cap BD$ $ ).$ Prove that $ PQ\parallel EF\parallel CD,$ where $ P\equiv AB\cap DE$ and $ Q\equiv CD\cap AF.$ PROOF. – We consider the triads of the points $ A,\ P,\ D$ and $ B,\ Q,\ C,$ on the line segments $ AD,\ BC$ respectively and applying the Pappos theorem, we have that the points $ X\equiv CP\cap DQ$ and $ Y\equiv AC\cap BD$ and $ Z\equiv AQ\cap BP,$ are collinear. From this collinearity, based on the Desarques theorem, we have that the triangles $ \bigtriangleup CEP$ and $ \bigtriangleup DFQ$ are perspective and so, we conclude that the line segments $ CD,\ EF,\ PQ$ are concurrent at infinity, because of $ EF\parallel CD.$ Hence, we conlude that $ PQ\parallel EF\parallel CD$ and the proof of the Lemma is completed. $ \bullet$ Return now in the configuration of the proposed problem, let $ EFKL$ be, the parallelogram as it has been stated in its formulation, with $ EF\parallel KL$ and $ EL\parallel FK.$ We consider the quadrilateral $ PQLK$ and the points $ E,\ F$ on $ QK,\ PL$ pespectively, of which $ EF\parallel KL.$ So, based on the above Lemma, we have that $ EF\parallel BD\parallel KL$ and similarly, $ EL\parallel AC\parallel FK.$ Applying now, the problem in the topic Locus of center parallelogram of N.T.TUAN, we conclude that the point $ O\equiv EK\cap FL,$ lies on the line segment $ MN,$ where $ M,\ N$ are the midpoints of the segments $ BD,\ AC$ respectively and the proof of the proposed problem, is completed. Kostas Vittas.

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t=279925(a).pdf (4kb)
t=279925.pdf (4kb)