Geometry Problem: Prove AB=BE=EA

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Homework Statement



http://img58.imageshack.us/img58/1977/49718826ce7.png

you need to prove that AB=BE=EA

Homework Equations


only that angleEDC = angleECD=15°
and ABCD is a square
and it should be solved used only geometry, with no trigo.


The Attempt at a Solution


nvm, got it
 
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http://geometri-problemleri.blogspot.com/2009/11/problem-52-ve-cozumu.html
 
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Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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