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Geometry Difficulty 4.5 AIME Prove it Romania

Consider ABCDABCD a rectangle of center OO with ABBCAB \ne BC. The perpendicular dropped from OO to BDBD intersects lines ABAB and BCBC in EE and FF. Let MM and NN be the midpoints of segments [CD][CD] and [AD][AD]. Prove that FMENFM \perp EN.

Solution

Let PP be the midpoint of [BC][BC] and QQ be the intersection point of EOEO and CDCD. As [PM][PM] is the midsegment of the triangle BCDBCD, it results that PMBDPM \parallel BD and OQBDOQ \perp BD, so QFPMQF \perp PM. But PCMQPC \perp MQ, so FF is the orthocenter of the triangle MPQMPQ, which leads to PQMFPQ \perp MF. Since the quadrilateral ENQPENQP is a parallelogram, we have PQENPQ \parallel EN. Consequently, FMENFM \perp EN.

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