The mid-point of the sides of triangle along with any of the vertices as the fourth point make a parallelogram of area equal to

Options

A. `(1)/(2) ar (ABC)`
B. `(1)/(3) ar (ABC)`
C. `(1)/(4) ar (ABC)`
D. `ar (ABC)`

Explanation

To solve the problem, we need to find the area of the parallelogram formed by the midpoints of the sides of triangle ABC and one of its vertices. Let's go through the solution step by step.

### Step-by-Step Solution:

1. **Identify the Triangle and Midpoints**:
Let triangle ABC have vertices A, B, and C. We denote the midpoints of sides AB, BC, and AC as D, E, and F respectively.

2. **Understanding the Parallelogram**:
We will consider the parallelogram formed by one vertex (say A) and the midpoints D, E, and F. For our case, we will focus on the parallelogram ADEF.

3. **Area of Triangle ABC**:
The area of triangle ABC can be denoted as Area(ABC).

4. **Area of Triangles Formed by Midpoints**:
Since D, E, and F are midpoints, the triangles ADF, DEF, and EBC are formed. The area of triangle DEF will be equal to one-fourth of the area of triangle ABC. This is because the midpoints divide the triangle into smaller triangles of equal area.

\[ \text{Area}(DEF) = \frac{1}{4} \text{Area}(ABC) \]

5. **Area of Triangle ADF**:
The area of triangle ADF is equal to the area of triangle DEF because they share the same base DF and height from A to line DF.

\[ \text{Area}(ADF) = \text{Area}(DEF) = \frac{1}{4} \text{Area}(ABC) \]

6. **Area of Parallelogram ADEF**:
The area of the parallelogram ADEF is equal to twice the area of triangle ADF (since a parallelogram can be split into two triangles).

\[ \text{Area}(ADEF) = 2 \times \text{Area}(ADF) = 2 \times \frac{1}{4} \text{Area}(ABC) = \frac{1}{2} \text{Area}(ABC) \]

7. **Conclusion**:
Therefore, the area of the parallelogram ADEF is equal to half the area of triangle ABC.

### Final Answer:
The area of the parallelogram formed by the midpoints of the sides of triangle ABC and one of its vertices is \( \frac{1}{2} \text{Area}(ABC) \).

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