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 $BM = EN$.
BdMO National 2012: Higher Secondary, Secondary 03
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
- nafistiham
- Posts:829
- Joined:Mon Oct 17, 2011 3:56 pm
- Location:24.758613,90.400161
- Contact:
Re: BdMO National 2012: Higher Secondary, Secondary 03
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.