Question

Prove that the diagonals of a rectangle bisect each other and are equal.

Answer 1

Concept:
Application:

Let OACB be a rectangle such that OA is along x-axis and OB is along y-axis. Let OA = a and OB = b.

Then, the coordinates of A and B are (a,0) and (0, b) respectively

Since, OACB is a rectangle. Therefore, $AC=OB\Rightarrow AC=b$

Thus, we have

OA = a and AC = b

So, the coordinates of C are (a, b)

The coordinates of the mid-points of OC are $(\frac{a+0}{2},\frac{b+0}{2})=\frac{a}{2},\frac{b}{2}$

Also, the coordinates of the mid-points of AB are $(\frac{a+0}{2},\frac{0+b}{2})=\frac{a}{2},\frac{b}{2}$

Clearly, coordinates of the mid-point of OC and AB are same.

Hence, OC and AB bisect each other

Also, $OC=\sqrt{a^2+b^2}$ and $AB=\sqrt{(a-0{)}^2+(0-b{)}^2}=\sqrt{a^2+b^2}$

$\therefore OC=AB$