D and E , are the mid-points of AB and AC of `Delta ABC, BC ` is produced to any point P. DE, DP and EP are joined, then `Delta PED` is equal to ?

Answer

To solve the problem, we need to analyze the triangle ABC and the points D and E, which are the midpoints of sides AB and AC, respectively. We will then consider the triangle formed by points P, D, and E.

### Step-by-Step Solution:

1. **Identify the Triangle and Midpoints**:
- Let triangle ABC be given with vertices A, B, and C.
- Identify D as the midpoint of AB and E as the midpoint of AC.

2. **Extend Side BC to Point P**:
- Extend line segment BC to a point P such that BC is produced to point P.

3. **Join Points D, E, and P**:
- Draw line segments DE, DP, and EP to form triangle PED.

4. **Use the Midpoint Theorem**:
- According to the midpoint theorem, DE is parallel to BC and DE is half the length of BC.
- This means that triangle ABC can be divided into smaller triangles with equal ratios.

5. **Area Relations**:
- The area of triangle ABC can be expressed in terms of the areas of triangles formed by the midpoints. Specifically, the area of triangle ABC is twice the area of triangle ADE (where D and E are midpoints).
- Since D and E are midpoints, triangle ADE is half the area of triangle ABC.

6. **Establish Area of Triangle PED**:
- Since DE is parallel to BC and D and E are midpoints, triangle PED is similar to triangle ABC.
- The area of triangle PED can be established as being equal to 1/4 of the area of triangle ABC because the height from point P to line DE is the same as the height from point A to line BC, and the base DE is half the length of BC.

7. **Conclusion**:
- Therefore, the area of triangle PED is equal to the area of triangle ABC divided by 4.

### Final Answer:
The area of triangle PED is equal to **(1/4) * Area of triangle ABC**.

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