Problem 1000

Let the diagonal $B E$ intersect the diagonals $A D$ and $A C$ at points $F$ and $G$. The sides of triangles $A F E$ and $B C D$ are parallel, so they are similar and $A F: F E = B C: C D$. Therefore, $A D: B E = (A F + B C):(E F + C D) = B C: C D$. Similarly, $A E: B D = D E: A C$. From the similarity of triangles $B E D$ and $E G A$, we get $A E: D B = E G: B E = C D: B E$. Thus, $\frac{B C}{A D} = \frac{C D}{B E} = \frac{A E}{B D} = \frac{D E}{A C} = \lambda$. Clearly, $\overrightarrow{B C} + \overrightarrow{C D} + \overrightarrow{D E} + \overrightarrow{E A} + \overrightarrow{A B} = \overrightarrow{0}, \overrightarrow{A D} + \overrightarrow{B E} + \overrightarrow{C A} +$

$\overrightarrow{D B} + \overrightarrow{E C} = \overrightarrow{0}$ and $\overrightarrow{B C} = \lambda \overrightarrow{A D}, \overrightarrow{C D} = \lambda \overrightarrow{B E}, \overrightarrow{D E} = \lambda \overrightarrow{C A}, \overrightarrow{E A} = \lambda \overrightarrow{D B}$. Therefore, $\overrightarrow{0} = \lambda(\overrightarrow{A D} + \overrightarrow{B E} + \overrightarrow{C A} + \overrightarrow{D B}) + \overrightarrow{A B} = -\lambda \overrightarrow{E C} + \overrightarrow{A B}$, i.e., $\overrightarrow{A B} = \lambda \overrightarrow{E C}$. Hence, $A B \parallel E C$.