Question

using analytic geometry to prove that the sum of the squares of the lengths of the sides of a parallelogram is equal to the sum of the squares of the lengths of the diagonals.

Explanation:

Step1: Assign Coordinates

Let the vertices of the parallelogram be \( A(0,0) \), \( B(a,0) \), \( C(a + b,c) \), \( D(b,c) \). This is because in a parallelogram, opposite sides are equal and parallel, so vector \( \overrightarrow{AB}=(a,0) \) and \( \overrightarrow{AD}=(b,c) \), so \( C = B+AD=(a + b,c) \).

Step2: Calculate Side Lengths Squared

Length of \( AB \): distance between \( A(0,0) \) and \( B(a,0) \). Using distance formula \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \), so \( AB^2=(a - 0)^2+(0 - 0)^2=a^2 \).
Length of \( BC \): distance between \( B(a,0) \) and \( C(a + b,c) \). \( BC^2=(a + b - a)^2+(c - 0)^2=b^2 + c^2 \).
Length of \( CD \): distance between \( C(a + b,c) \) and \( D(b,c) \). \( CD^2=(b-(a + b))^2+(c - c)^2=(-a)^2=a^2 \).
Length of \( DA \): distance between \( D(b,c) \) and \( A(0,0) \). \( DA^2=(b - 0)^2+(c - 0)^2=b^2 + c^2 \).
Sum of squares of sides: \( AB^2+BC^2+CD^2+DA^2=a^2+(b^2 + c^2)+a^2+(b^2 + c^2)=2a^2 + 2b^2+2c^2 \).

Step3: Calculate Diagonal Lengths Squared

Length of diagonal \( AC \): distance between \( A(0,0) \) and \( C(a + b,c) \). \( AC^2=(a + b - 0)^2+(c - 0)^2=(a + b)^2+c^2=a^2 + 2ab + b^2 + c^2 \).
Length of diagonal \( BD \): distance between \( B(a,0) \) and \( D(b,c) \). \( BD^2=(b - a)^2+(c - 0)^2=(b - a)^2+c^2=a^2 - 2ab + b^2 + c^2 \).
Sum of squares of diagonals: \( AC^2+BD^2=(a^2 + 2ab + b^2 + c^2)+(a^2 - 2ab + b^2 + c^2)=2a^2 + 2b^2+2c^2 \).

Step4: Compare Sums

Since sum of squares of sides \( 2a^2 + 2b^2+2c^2 \) is equal to sum of squares of diagonals \( 2a^2 + 2b^2+2c^2 \), the statement is proved.

Answer:

The sum of the squares of the lengths of the sides of a parallelogram is equal to the sum of the squares of the lengths of the diagonals (proven using analytic geometry as shown above).