BdMO National 2012: Higher Secondary, Secondary 03

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Moon
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BdMO National 2012: Higher Secondary, Secondary 03

Unread post by Moon » Sat Feb 11, 2012 11:19 pm

Problem:
In a given pentagon $ABCDE$, triangles $ABC, BCD, CDE, DEA$ and $EAB$ all have the same area. The lines $AC$ and $AD$ intersect $BE$ at points $M$ and $N$. Prove that $BM = EN$.
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Re: BdMO National 2012: Higher Secondary, Secondary 03

Unread post by nafistiham » Fri Feb 17, 2012 3:10 pm

sec 3 2012.JPG
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here, $\triangle BCD$ and $\triangle CDE$ are of the same area.so,$CD||BE$
likewise,for $\triangle BCD$ and $\triangle ABC$, $BC||AD$
for $\triangle CDE$ and $\triangle DEA$ $DE||CA$
from this we can say that $BNDC$ and $MCDE$ are parallelograms
now, $BN=CD=ME$
so, $BM=EN$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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