40th Hellenic Mathematical Olympiad

Since $ME\parallel \Gamma \Delta$ it follows that $\Gamma \Delta E=M\hat{E}\Delta$, and so $$180^{\circ }-\Gamma \Delta E=180^{\circ }-M\hat{E}\Delta \Rightarrow B\Delta \Gamma =M\hat{E}N(1)$$ Since the points $M$, $N$ are the midpoints of the sides $AB$, $A\Gamma$, respectively, we conclude that $$MN\parallel B\Gamma ,MN=\frac{B\Gamma }{2}.(2)$$ and hence $$\Delta B\hat{\Gamma }=M\hat{N}E,(3)$$ From (1) and (3) we get that the triangles $B\Delta \Gamma$ and $MEN$ are similar, and so: $$\frac{B\Delta }{EN}=\frac{B\Gamma }{MN}=2\Rightarrow B\Delta =2\cdot EN.$$

Figure 1

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Alternative solution.

From $\Gamma$ we draw the parallel to the line to $BN$, which meet the line $AE$ at $Z$. Since $N$ is the midpoint of $B\Gamma$ and $NE\parallel \Gamma Z$, it follows that $E$ is the midpoint of $AZ$ and $\Gamma Z=2\cdot EN$. Since $M$, $E$ are the midpoints of $AB$, $AZ$, respectively, we have: $ME\parallel B\Gamma$. Therefore the quadrilateral $BZ\Gamma \Delta$ is parallelogram, as it has the two pairs of opposite sides parallel. Hence $B\Delta =\Gamma Z=2\cdot EN$.