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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 932.
This problem is similar to problem 879. refer to links below for solutionhttp://gogeometry.blogspot.com/2013/05/problem-879-square-midpoint-diagonal.htmlhttp://img208.imageshack.us/img208/6976/problem897.pngangle(FED)=angle(EDF)= 45 so angle(EFD)=90
As shown by Peter, it's easy to show that FE=FD and ///ly FB=FE. Thus, a circle drawn with F as the centre and radius=FE passes through points D & B as well. Therefore, /_DFE=2*/_DBE =90°
A graphical solution: http://bleaug.free.fr/gogeometry/932.pngbleaug
Let A(0,0), B (0,4a), C(4a,4a), D(4a,0) be the vertices.Clearly E(2a,4a) and F(a,a) are the other points.EF^2 = 10a^2 = DF^2, ED^2 = 20a^2.ED^2 = EF^2+ DF^2.By Pythagoras EFD is a right angle.