Square

**james_bond** · Apr 8th 2010, 10:10 AM

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 $EDF\angle =45^\circ$ .

**running-gag** · Apr 8th 2010, 12:08 PM

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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