prove that in a parallelogram, the sum of the squares of the lengths of the diagonals equals the sum of the squares of the lengths of the four sides

We are tasked with proving that in a parallelogram, the sum of the squares of the diagonals equals the sum of the squares of the four sides. This is a classical result from Euclidean geometry and can be derived using the **parallelogram law** and vector algebra.

### Let's outline the steps to prove this:

Let the parallelogram be defined by vertices \( A \), \( B \), \( C \), and \( D \), where \( A \) and \( C \) are opposite vertices, and \( B \) and \( D \) are opposite vertices.

Denote the lengths of the sides as \( AB = CD = b \) and \( AD = BC = a \), and let \( \vec{A} = \vec{0} \), \( \vec{B} = \vec{u} \), \( \vec{D} = \vec{v} \), and \( \vec{C} = \vec{u} + \vec{v} \), where \( \vec{u} \) and \( \vec{v} \) are vectors representing sides of the parallelogram.

The diagonals of the parallelogram are \( AC \) and \( BD \). Using vectors:
- Diagonal \( AC \) is given by \( \vec{AC} = \vec{C} - \vec{A} = \vec{u} + \vec{v} \)
- Diagonal \( BD \) is given by \( \vec{BD} = \vec{B} - \vec{D} = \vec{u} - \vec{v} \)

### Step 1: Compute the squares of the diagonals
The squares of the lengths of diagonals are the dot products:
\[
AC^2 = |\vec{u} + \vec{v}|^2 = (\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v}) = \vec{u} \cdot \vec{u} + 2 \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{v}
\]
\[
BD^2 = |\vec{u} - \vec{v}|^2 = (\vec{u} - \vec{v}) \cdot (\vec{u} - \vec{v}) = \vec{u} \cdot \vec{u} - 2 \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{v}
\]

### Step 2: Sum of the squares of the diagonals
Adding the two expressions for the diagonals:
\[
AC^2 + BD^2 = (\vec{u} \cdot \vec{u} + 2 \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{v}) + (\vec{u} \cdot \vec{u} - 2 \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{v})
\]
\[
AC^2 + BD^2 = 2(\vec{u} \cdot \vec{u} + \vec{v} \cdot \vec{v})
\]
Since \( \vec{u} \cdot \vec{u} = a^2 \) (the length of side \( AD \)) and \( \vec{v} \cdot \vec{v} = b^2 \) (the length of side \( AB \)), we get:
\[
AC^2 + BD^2 = 2(a^2 + b^2)
\]

### Step 3: Sum of the squares of the sides
The sum of the squares of the four sides of the parallelogram is:
\[
2a^2 + 2b^2
\]

### Conclusion:
We have shown that:
\[
AC^2 + BD^2 = 2a^2 + 2b^2
\]
Thus, the sum of the squares of the diagonals equals the sum of the squares of the four sides, which completes the proof.

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Would you like more details on any part of the proof or further clarification?

### Related Questions:
1. What are the properties of the diagonals in a parallelogram?
2. How can we generalize this proof to other quadrilaterals like rectangles?
3. How does this result relate to the parallelogram law in physics?
4. Can we derive this result using coordinate geometry?
5. How does this relate to the law of cosines for parallelograms?

### Tip:
When working with vectors, always remember that the dot product \( \vec{u} \cdot \vec{v} \) gives a powerful way to relate angles and lengths algebraically.

Prove that in a parallelogram, the sum of the squares of the lengths of the diagonals equals the sum of the squares of the lengths of the four sides.

Learn how to prove that in a parallelogram, the sum of the squares of the diagonals equals the sum of the squares of the four sides. This detailed solution uses vector algebra and the parallelogram law. Suitable for high school students in Grades 10-12.

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