Chapter 6.2, Problem 59E

Solution

To prove: that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Given information:

Given, u is any vector.

Calculation:

Consider ABCD be a parallelogram and its diagonals AC and BD intersect at the point O.

  

In triangle ABC, BO is the median of the diagonal BD since diagonals of a parallelogram bisect each other.

Then,

  $A{B}^2+B{C}^2=\left(O{A}^2+O{B}^2\right)+\left(O{B}^2+O{C}^2\right)$

Since,

  $OC=OA$

Then,

  $\begin{array}{c}A{B}^2+B{C}^2=\left(O{A}^2+O{B}^2\right)+\left(O{B}^2+O{A}^2\right)\\ =2O{A}^2+2O{B}^2\end{array}$

In triangle ADC, DO is the median of the diagonal BD .

Then,

  $\begin{array}{c}A{D}^2+D{C}^2=\left(A{O}^2+O{D}^2\right)+\left(O{D}^2+O{C}^2\right)\\ =\left(A{O}^2+O{D}^2\right)+\left(O{D}^2+O{A}^2\right)\\ =2A{O}^2+2O{D}^2\end{array}$

On adding both these expressions in below.

  $A{B}^2+B{C}^2+A{D}^2+D{C}^2=\left(2O{A}^2+2O{B}^2\right)+\left(2A{O}^2+2O{D}^2\right)$

Since,

  $OB=OD$

Then,

  $\begin{array}{c}A{B}^2+B{C}^2+A{D}^2+D{C}^2=\left(2O{A}^2+2O{B}^2\right)+\left(2A{O}^2+2O{B}^2\right)\\ =4\left[O{A}^2+O{B}^2\right]\end{array}$

Also, $OA=\frac{AC}{2}$ and $OB=\frac{BD}{2}$ .

So,

  $\begin{array}{c}A{B}^2+B{C}^2+A{D}^2+D{C}^2=4\left[{\left(\frac{AC}{2}\right)}^2+{\left(\frac{BD}{2}\right)}^2\right]\\ =4\times \frac{1}{4}\left[A{C}^2+B{D}^2\right]\\ =A{C}^2+B{D}^2\end{array}$

Hence,

  $A{B}^2+B{C}^2+A{D}^2+D{C}^2=A{C}^2+B{D}^2$