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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 731 details.
Note: Please draw a figure, as indicated.Let O be the center of the circle.Let M be the midpoint of AD.Let G be the centroid of triangle BAD. Let BM be the median of triangle BAD, joining the vertex B to M.Extend OG to E'such that GE' = 2OGConsider triangles BE'G and GOM.BG:GM = 2:1 (Property of median)GE':OG = 2:1 (construction of E')So BG:GM = GE':OGAlso BE' is parallel to OM=> BE' is perpendicular to AD.=> E' lies on the altitude from B (on AD)By a similar argument E' lies on the altitude from A on the side BD (of triangle BAD.So E' is the orthocenter of triangle BAD.Hence E' coincides with E.By proportionality of corresponding sides in the similar triangles BEG and OMG,BE:OG = BG:GM = 2:1Thus BE = 2 OM.Noting that same O is the circumcenter and F is the orthocenter of triangle ACD, it follows by an exactly similar argument that CF = 2OMHence BE = CF. Also BE is parallel to CF (each being perpendicular to AD)Hence BEFC is a parallelogram.
http://img715.imageshack.us/img715/3812/problem731.pngBE and CF cut the circle at E’ and F’Note that AD is the perpen. Bisector of EE’ and FF’ ( property of orthocenters)And EFF’E’ and BCF’E’ are isosceles trapezoidsAnd angle(BE’F’)=angle(E’EF)=(E’BC) => BC//EF => BCFE is a parallelogram
Problem 731Ιf O is the center of the circle ABCD and OM perpendicular AD .Τhen BE=//2.OMCF=//2.OM. So BE=//CF therefore BCFE is parallelogram.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE