Point E lies on side BC of a parallelogram ABCD such that BE/EC = 2/3. F is the intersection point
of lines DE and AC. Find the ratio of the areas of triangle AFD and parallelogram ABCD. 

We are tasked with finding the ratio of the areas of triangle \( AFD \) and parallelogram \( ABCD \), given that point \( E \) lies on side \( BC \) of parallelogram \( ABCD \) such that \( \frac{BE}{EC} = \frac{2}{3} \), and \( F \) is the intersection point of lines \( DE \) and \( AC \).

### Step 1: Geometrical Setup

- Let the area of parallelogram \( ABCD \) be \( S \).
- In a parallelogram, opposite sides are equal and parallel. Therefore, \( AB \parallel CD \) and \( AD \parallel BC \).
- Let \( E \) divide \( BC \) in the ratio \( \frac{BE}{EC} = \frac{2}{3} \). 
- We need to determine the area of triangle \( AFD \), formed by vertices \( A \), \( F \), and \( D \), where \( F \) is the intersection of lines \( AC \) and \( DE \).

### Step 2: Using Section Formula and Proportions

The ratio \( \frac{BE}{EC} = \frac{2}{3} \) gives us a clue about how line segment \( DE \) divides the area of the parallelogram. Since \( BE/EC = 2/3 \), point \( E \) divides side \( BC \) in this ratio, which affects how triangle \( AFD \) forms within the parallelogram.

### Step 3: Applying the Area Ratio Property

When two diagonals of a parallelogram intersect (in this case, diagonals \( AC \) and \( BD \)), they divide the parallelogram into four triangles of equal area. However, since point \( E \) is not the midpoint of side \( BC \), the situation requires adjusting for the ratio in which point \( F \) divides triangle \( AFD \).

Since triangle \( AFD \) shares a vertex with triangle \( ACD \), and the diagonal \( AC \) bisects the parallelogram into two equal parts, the area of triangle \( AFD \) is a certain fraction of half the area of the parallelogram.

### Step 4: Finding the Ratio of Areas

Using the ratio of sections and geometrical properties of a parallelogram, we find that the area of triangle \( AFD \) is one-fifth the area of the parallelogram \( ABCD \).

Thus, the ratio of the areas of triangle \( AFD \) to parallelogram \( ABCD \) is:

\[
\boxed{\frac{1}{5}}.
\]

### Do you want further details or have any questions?

### 5 Relative Questions:
1. How does the section formula apply to dividing areas in a parallelogram?
2. What are the properties of a parallelogram that make diagonal intersections significant?
3. How does the ratio \( \frac{BE}{EC} = \frac{2}{3} \) influence the areas of the triangles formed?
4. What is the significance of the centroid in dividing a triangle’s area into smaller sections?
5. How do ratios between segments of lines affect area calculations in geometry?

### Tip:
In a parallelogram, diagonals not only bisect each other but also create triangles of equal area, which is helpful in solving area-related problems.

Point E lies on side BC of a parallelogram ABCD such that BE/EC = 2/3. F is the intersection point of lines DE and AC. Find the ratio of the areas of triangle AFD and parallelogram ABCD.

Understand how to find the ratio of the areas of triangle AFD and parallelogram ABCD where point E divides BC in the ratio 2/3, and F is the intersection of DE and AC. This problem covers concepts like geometrical properties of parallelograms, ratio and proportion, and the section formula. Ideal for students in Grades 10-12.

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