Let $A,B,C$ be points on $[OX$ and $D,E,F$ be points on $[OY$ such that $|OA|=|AB|=|BC|$ and $|OD|=|DE|=|EF|$. If $|OA|>|OD|$, which one below is true?

$\textbf{(A)}$ For every $\widehat{XOY}$, $\text{ Area}(AEC)>\text{Area}(DBF)$

$\textbf{(B)}$ For every $\widehat{XOY}$, $\text{ Area}(AEC)=\text{Area}(DBF)$

$\textbf{(C)}$ For every $\widehat{XOY}$, $\text{ Area}(AEC)<\text{Area}(DBF)$

$\textbf{(D)}$ If $m(\widehat{XOY})<45^\circ$ then $\text{Area}(AEC)<\text{Area}(DBF)$, and if $45^\circ < m(\widehat{XOY})<90^\circ$  then $\text{Area}(AEC)>\text{Area}(DBF)$

$\textbf{(E)}$ None of above
1. Let \( OA = a \) and \( OD = b \). Given that \( |OA| = |AB| = |BC| \) and \( |OD| = |DE| = |EF| \), we can express the coordinates of points \( A, B, C \) and \( D, E, F \) as follows:
   - \( A \) is at \( a \) on \( [OX \)
   - \( B \) is at \( 2a \) on \( [OX \)
   - \( C \) is at \( 3a \) on \( [OX \)
   - \( D \) is at \( b \) on \( [OY \)
   - \( E \) is at \( 2b \) on \( [OY \)
   - \( F \) is at \( 3b \) on \( [OY \)

2. We need to find the areas of triangles \( AEC \) and \( DBF \). The area of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
   \[
   \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
   \]

3. For triangle \( AEC \):
   - \( A \) is at \( (a, 0) \)
   - \( E \) is at \( (0, 2b) \)
   - \( C \) is at \( (3a, 0) \)
   Using the area formula:
   \[
   \text{Area}(AEC) = \frac{1}{2} \left| a(2b - 0) + 0(0 - 0) + 3a(0 - 2b) \right| = \frac{1}{2} \left| 2ab - 6ab \right| = \frac{1}{2} \left| -4ab \right| = 2ab
   \]

4. For triangle \( DBF \):
   - \( D \) is at \( (0, b) \)
   - \( B \) is at \( (2a, 0) \)
   - \( F \) is at \( (0, 3b) \)
   Using the area formula:
   \[
   \text{Area}(DBF) = \frac{1}{2} \left| 0(0 - 3b) + 2a(3b - b) + 0(b - 0) \right| = \frac{1}{2} \left| 0 + 2a \cdot 2b + 0 \right| = \frac{1}{2} \left| 4ab \right| = 2ab
   \]

5. Since \( \text{Area}(AEC) = 2ab \) and \( \text{Area}(DBF) = 2ab \), it follows that for every \( \angle XOY \), the areas of triangles \( AEC \) and \( DBF \) are equal.

Conclusion:
\[
\boxed{\textbf{(B)}}
\]