Hellenic Mathematical Olympiad

Since $ME \parallel ΓΔ$ it follows that $ΓΔE = M\hat{E}Δ$, and so
$$180^\circ - ΓΔE = 180^\circ - M\hat{E}Δ \Rightarrow BΔΓ = M\hat{E}N \quad (1)$$
Since the points $M$, $N$ are the midpoints of the sides $AB$, $AΓ$, respectively, we conclude that
$$MN \parallel BΓ, \quad MN = \frac{BΓ}{2}. \quad (2)$$
and hence
$$ΔB\hat{Γ} = M\hat{N}E, \quad (3)$$
From (1) and (3) we get that the triangles $BΔΓ$ and $MEN$ are similar, and so:
$$\frac{BΔ}{EN} = \frac{BΓ}{MN} = 2 \Rightarrow BΔ = 2 \cdot EN.$$

Figure 1

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