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by Corei13 » Mon Jan 17, 2011 11:21 pm
it can be done using only pythagorus's theorem and "segment joing midpoints of two sides of a triangle is half of the third side :-/
Let, $AB=a$
$\Longrightarrow AN^2 = \frac{5a^2}{4}$
and,
$AM^2 = AO^2 + OM^2 = {\frac{a}{\sqrt{2}}}^2 + {\frac{a}{2\sqrt{2}}}^2 = \frac{5a^2}{8}$
Let, $MZ || BC$ and $MZ$ intersect $CD$ on $Z$, and $AB$ on $Y$
clearly $NZ = \frac{a}{4}$, $MZ = a-MY = \frac{3a}{4}$
then, $MN^2 = MZ^2 + NZ^2 = \frac{5a^2}{8}$
So, $AM=MN$, and $AN^2 = AM^2 + MN^2$, Done!
Last edited by
Corei13 on Tue Jan 18, 2011 5:32 pm, edited 1 time in total.
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