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The problem appears to involve a parallelogram \(ABCD\) with specific points and relationships, as well as various geometric questions. Let's break down and solve it step by step:

### Problem Description:
1. **Given**: 
    - \(ABCD\) is a parallelogram.
    - Point \(E\) is on segment \(BC\).
    - \( \frac{BE}{CE} = \frac{2}{3} \).
    - The area of triangle \(GEC\) is 18 square units.
2. **Tasks**:
    1. Prove that \( \Delta AEB \sim \Delta GEC \).
    2. Calculate the area of triangle \(AEB\).
    3. Prove that \( \Delta AGD \sim \Delta GEC \).
    4. Calculate the area of triangle \(GAD\).
    5. Calculate the area of parallelogram \(ABCD\).

### Step 1: Prove \( \Delta AEB \sim \Delta GEC \)

- **Triangles** \( \Delta AEB \) and \( \Delta GEC \) can be proven similar using the **AA (Angle-Angle) similarity criterion**.
- Since \(ABCD\) is a parallelogram, \(AB \parallel CD\) and \(AD \parallel BC\). Therefore, angles \( \angle AEB \) and \( \angle GEC \) are equal because corresponding angles between parallel lines are equal.
- Also, \( \angle BAE = \angle CGE \) because opposite angles of a parallelogram are equal.
- Therefore, \( \Delta AEB \sim \Delta GEC \).

### Step 2: Calculate the area of triangle \(AEB\)

- Since \( \Delta AEB \sim \Delta GEC \) and their similarity ratio is the same as \( \frac{BE}{CE} = \frac{2}{3} \), the ratio of their areas is the square of their corresponding sides.
- Let the area of \( \Delta AEB \) be \( x \). Then:
  \[
  \frac{x}{18} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}
  \]
  Solving for \( x \):
  \[
  x = \frac{4}{9} \times 18 = 8 \text{ square units}
  \]

### Step 3: Prove \( \Delta AGD \sim \Delta GEC \)

- The triangles \( \Delta AGD \) and \( \Delta GEC \) can be proven similar using similar logic to Step 1.
- Angle \( \angle AGD = \angle GEC \) (as they are corresponding angles).
- \( \angle GAD = \angle GCE \) (opposite angles in a parallelogram are equal).
- Hence, \( \Delta AGD \sim \Delta GEC \).

### Step 4: Calculate the area of triangle \(GAD\)

- The area of similar triangles is proportional to the square of their corresponding sides.
- Since the triangles \( \Delta AGD \) and \( \Delta GEC \) are similar, the area of \( \Delta AGD \) is:
  \[
  \text{Area of } \Delta AGD = \frac{ \text{Area of } \Delta GEC }{\left(\frac{BE}{CE}\right)^2} = \frac{18}{\frac{4}{9}} = 40.5 \text{ square units}
  \]

### Step 5: Calculate the area of parallelogram \(ABCD\)

- The area of parallelogram \(ABCD\) is twice the area of \( \Delta ABD \) (or any diagonal divides the parallelogram into two congruent triangles).
- The area of \( \Delta ABD = \) Area of \( \Delta AEB \) + Area of \( \Delta GAD \).
- Area of \(ABCD = 2 \times (8 + 40.5) = 97 \text{ square units}\).

---

### Summary:
- **Area of Triangle \(AEB\)**: 8 square units
- **Area of Triangle \(GAD\)**: 40.5 square units
- **Area of Parallelogram \(ABCD\)**: 97 square units

Would you like more details on any of these steps or have any other questions?

### Related Questions:
1. What is the relationship between the side ratios of similar triangles?
2. How can you find the area of a parallelogram using the area of a triangle within it?
3. What other properties can be inferred if a triangle is similar to another triangle?
4. How do you find the ratio of areas between two similar figures?
5. What geometric properties are preserved under similarity transformations?

### Tip:
When working with similar triangles, always remember that the ratio of their areas is the square of the ratio of their corresponding sides.

Explore a geometry problem involving parallelogram ABCD, triangles AEB, GEC, and AGD. Learn about triangle similarity, area calculations, and properties of parallelograms. Detailed step-by-step solution included.

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