**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 $BM = EN$.

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 $BM = EN$.

"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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- nafistiham
**Posts:**829**Joined:**Mon Oct 17, 2011 3:56 pm**Location:**24.758613,90.400161-
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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, $BM=EN$

\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]

Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.