Chapter 9.1, Problem 12P

To determine

To prove: 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 its diagonals.

Explanation of Solution

Given information: The points are $A\left(0,0\right),B\left(5,0\right),C\left(a+b,c\right)$ and $D\left(b,c\right)$ .

Formula used: $D=\sqrt{{\left(c-a\right)}^2+{\left(d-b\right)}^2}$

Proof: Apply the formula to sides lengths between the given points,

  $\begin{array}{l}\Rightarrow AB=\sqrt{{\left(0-a\right)}^2+{\left(0-0\right)}^2}\\ =a\\ \Rightarrow BE=\sqrt{{\left(a-\left(a+b\right)\right)}^2+{\left(0-c\right)}^2}\\ =\sqrt{b^2+c^2}\end{array}$

Similarly,

  $\begin{array}{l}\Rightarrow CD=\sqrt{{\left(a+b-b\right)}^2+{\left(c-c\right)}^2}\\ =a\\ \Rightarrow FC=\sqrt{{\left(b-0\right)}^2+{\left(c-0\right)}^2}\\ =\sqrt{b^2+c^2}\end{array}$

Hence, the sum of the square of the sides,

  $a^2+b^2+c^2+a^2+b^2+c^2=2a^2+2b^2+2c^2\mathrm{...}\left(1\right)$

Now, calculate the diagonals length,

  $\begin{array}{l}\Rightarrow AC=\sqrt{{\left(0-\left(a+b\right)\right)}^2+{\left(0-c\right)}^2}\\ =\sqrt{a^2+b^2+c^2+2ab}\\ \Rightarrow BD=\sqrt{{\left(a-b\right)}^2+{\left(0-c\right)}^2}\\ =\sqrt{a^2+b^2+c^2-2ab}\end{array}$

Hence, the sum of the squares of the diagonals,

  $a^2+b^2+c^2+2ab+a^2+b^2+c^2-2ab=2a^2+2b^2+2c^2\mathrm{...}\left(2\right)$

Since $2a^2+2b^2+2c^2=2a^2+2b^2+2c^2$

Hence, the sum of the lengths of the sides of a parallelogram is equal to the sum of the squares of the lengths of its diagonals.

Hence, proved.