Prove that the diagonals of a rhombus bisect each other at right angles using coordinate geometry.

Let the coordinates of the vertices be O(0, 0), A(a, 0), B(a + c, d) and C(c, d).

 Because it is a rhombus, all the sides are of equal length.

$OA=AB=BC=CO$

Gradient of $OB=\frac{d}{a+c}$and

Gradient of $AC=\frac{d}{c-a}$

The product of the gradients of the diagonals $=\frac{d}{a+c}\times \frac{d}{c-a}=\frac{d^2}{c^2-a^2}$

$O{A}^2=a^2$and by Pythagoras's theorem, $A{B}^2=c^2+d^2$

$a^2=c^2+d^2$

Hence $d^2=-\left(c^2-a^2\right)$

Thus, the product of the gradient of the diagonals $=-1.$