Given a convex pentagon $ABCDE$ with $AB=BC, CD=DE, \angle ABC=150^o, \angle CDE=30^o, BD=2$. Find the area of $ABCDE$.

(I.Voronovich)
1. **Identify the given information and set up the problem:**
   - We have a convex pentagon \(ABCDE\) with the following properties:
     - \(AB = BC\)
     - \(CD = DE\)
     - \(\angle ABC = 150^\circ\)
     - \(\angle CDE = 30^\circ\)
     - \(BD = 2\)

2. **Understand the transformations:**
   - The notation \(D_{30^\circ}(E) = C\) means that point \(E\) is rotated by \(30^\circ\) around point \(D\) to get point \(C\).
   - Similarly, \(B_{150^\circ}(C) = A\) means that point \(C\) is rotated by \(150^\circ\) around point \(B\) to get point \(A\).

3. **Combine the transformations:**
   - The combined transformation \(D_{30^\circ} \circ B_{150^\circ}(E) = A\) implies that point \(E\) is first rotated by \(150^\circ\) around point \(B\) and then by \(30^\circ\) around point \(D\) to get point \(A\).
   - Since \(30^\circ + 150^\circ = 180^\circ\), the combined transformation is equivalent to a rotation by \(180^\circ\), which is a reflection over the midpoint of \(AE\).

4. **Determine the midpoint \(M\) of \(AE\):**
   - Since the combined transformation is a reflection over the midpoint \(M\) of \(AE\), we have \(\angle MDB = 15^\circ\) and \(\angle DBM = 75^\circ\).
   - Therefore, \(\angle DMB = 90^\circ\).

5. **Construct the isosceles triangle \(BDF\):**
   - Let \(F\) be a point on \(BM\) such that \(FM = MB\) and \(F \neq B\).
   - The triangle \(BDF\) is isosceles with \(DF = DB = 2\).

6. **Calculate the area of the pentagon \(ABCDE\):**
   - The area of the pentagon can be decomposed into the areas of triangles \(\Delta ABM\), \(\Delta BMD\), and \(\Delta CDE\).
   - Since \(\Delta BMA \cong \Delta MEF\), we have:
     \[
     \text{Area}(ABCDE) = \text{Area}(ABM) + \text{Area}(BMD) + \text{Area}(CDE)
     \]
   - Given that \(\Delta BDF\) is isosceles and \(\angle DMB = 90^\circ\), the area of \(\Delta BDF\) is:
     \[
     \text{Area}(BDF) = \frac{1}{2} \cdot BD \cdot DF \cdot \sin(30^\circ) = \frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{1}{2} = 1
     \]

The final answer is \(\boxed{1}\).