State and prove the parallelogram law.
Understand the Problem
The question asks for the statement and proof of the parallelogram law, which is a fundamental result in Euclidean geometry relating the lengths of the sides and diagonals of a parallelogram.
Answer
$2a^2 + 2b^2 = d_1^2 + d_2^2$
Answer for screen readers
The parallelogram law states that for a parallelogram with sides of length $a$ and $b$, and diagonals of length $d_1$ and $d_2$, the following equation holds: $$2a^2 + 2b^2 = d_1^2 + d_2^2$$ Proof: Let $\vec{AB} = \vec{u}$ and $\vec{AD} = \vec{v}$. Then $d_1^2 = |\vec{u} + \vec{v}|^2 = a^2 + 2\vec{u} \cdot \vec{v} + b^2$ and $d_2^2 = |\vec{v} - \vec{u}|^2 = a^2 - 2\vec{u} \cdot \vec{v} + b^2$. Adding these gives $d_1^2 + d_2^2 = 2a^2 + 2b^2$.
Steps to Solve
- State the Parallelogram Law
The parallelogram law states that the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals. Let the parallelogram be $ABCD$, where $AB = CD = a$ and $BC = DA = b$. Let the diagonals be $AC = d_1$ and $BD = d_2$. Then the parallelogram law can be written as:
$$2a^2 + 2b^2 = d_1^2 + d_2^2$$
- Provide a Vector Proof
Let $\vec{AB} = \vec{u}$ and $\vec{AD} = \vec{v}$. Then $|\vec{u}| = a$ and $|\vec{v}| = b$. The diagonals can be represented as $\vec{AC} = \vec{u} + \vec{v}$ and $\vec{BD} = \vec{v} - \vec{u}$. The lengths of the diagonals squared are:
$$d_1^2 = |\vec{u} + \vec{v}|^2 = (\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v}) = |\vec{u}|^2 + 2\vec{u} \cdot \vec{v} + |\vec{v}|^2 = a^2 + 2\vec{u} \cdot \vec{v} + b^2$$
$$d_2^2 = |\vec{v} - \vec{u}|^2 = (\vec{v} - \vec{u}) \cdot (\vec{v} - \vec{u}) = |\vec{v}|^2 - 2\vec{u} \cdot \vec{v} + |\vec{u}|^2 = b^2 - 2\vec{u} \cdot \vec{v} + a^2$$
- Sum the Squares of the Diagonals
Add the two equations:
$$d_1^2 + d_2^2 = (a^2 + 2\vec{u} \cdot \vec{v} + b^2) + (a^2 - 2\vec{u} \cdot \vec{v} + b^2) = 2a^2 + 2b^2$$
Thus, $d_1^2 + d_2^2 = 2a^2 + 2b^2$, which proves the parallelogram law.
The parallelogram law states that for a parallelogram with sides of length $a$ and $b$, and diagonals of length $d_1$ and $d_2$, the following equation holds: $$2a^2 + 2b^2 = d_1^2 + d_2^2$$ Proof: Let $\vec{AB} = \vec{u}$ and $\vec{AD} = \vec{v}$. Then $d_1^2 = |\vec{u} + \vec{v}|^2 = a^2 + 2\vec{u} \cdot \vec{v} + b^2$ and $d_2^2 = |\vec{v} - \vec{u}|^2 = a^2 - 2\vec{u} \cdot \vec{v} + b^2$. Adding these gives $d_1^2 + d_2^2 = 2a^2 + 2b^2$.
More Information
The parallelogram law is a special case of Apollonius' theorem. It is also closely related to the polarization identity.
Tips
A common mistake is to confuse the parallelogram law with the Pythagorean theorem. The parallelogram law applies to parallelograms, while the Pythagorean theorem applies to right triangles. Another common mistake is errors in the vector algebra manipulations, especially with the dot product.
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