Chapter 13, Problem 20CT

To determine

To prove: The diagonals of a rectangle bisect each other.

Explanation of Solution

Proof:

Consider a rectangle $\text{ABCD}$ and the diagonals of rectangle intersect at P.

  

Figure (1)

To show that the diagonals AB and OC of the rectangle bisect each other, prove that $\text{OP}=\text{PC}$ and $\text{AP}=\text{PB}$ .

Let P is the midpoint of OC. The coordinates of point P can be calculated as:

  $\left(\frac{0+a}{2},\frac{0+b}{2}\right)=\left(\frac{a}{2},\frac{b}{2}\right)$

Now, the length of OP is:

  $\begin{array}{c}\text{OP}=\sqrt{{\left(\frac{a}{2}-0\right)}^2+{\left(\frac{b}{2}-0\right)}^2}\\ =\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}\end{array}$

The length of PC is:

  $\begin{array}{c}\text{PC}=\sqrt{{\left(a-\frac{a}{2}\right)}^2+{\left(b-\frac{b}{2}\right)}^2}\\ =\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}\end{array}$

Thus, $\text{OP}=\text{PC}$ . Similarly, $\text{AP}=\text{PB}$ .

Hence, it is proved that diagonals of a rectangle bisect each other.