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Math Help - Square

  1. #1
    Senior Member
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    Nov 2007
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    Square

    There is an ABCD square with side length of 1. E\in\overline{AB} and F\in\overline{BC} so that the perimeter of triangle EBF is 2. Show that EDF\angle =45^\circ.
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  2. #2
    MHF Contributor
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    Nov 2008
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    Hi

    I have found a solution but don't know if it is the simplest one !

    Let x=BE and y=BF
    We know that x+y+\sqrt{x^2+y^2} = 2 (perimeter of triangle EBF)

    On one hand
    \vec{DE}.\vec{DF} = \sqrt{1+(1-x)^2} \times \sqrt{1+(1-y)^2} \times \cos(\hat{EDF})

    On the other hand
    \vec{DE}.\vec{DF} = (\vec{DA}+\vec{AE}).(\vec{DC}+\vec{CF}) = (1-y)+(1-x) = 2-x-y

    Combining the 2 latter relations and using x+y+\sqrt{x^2+y^2} = 2 you can show that \cos^2(\hat{EDF}) = \frac{1}{2}
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