Cyclic quadrilaterals

A quadrilateral whose four vertices lie on one circle is called cyclic¹. We examine the one with vertices $A(5, 0)$ , $B(3, 4)$ , $C(-3, 4)$ and $D(-4, -3)$ .

They lie on one circle

All four points satisfy $x^2 + y^2 = 25$ .

Vertex	$x^2 + y^2$
$A(5, 0)$	$25 + 0 = 25$
$B(3, 4)$	$9 + 16 = 25$
$C(-3, 4)$	$9 + 16 = 25$
$D(-4, -3)$	$16 + 9 = 25$

They lie on the circle of center $(0, 0)$ and radius $5$ .

Opposite angles sum to a straight angle

In such a quadrilateral opposite angles add to $180^\circ$ . By the inscribed angle theorem, $\angle A$ is half the central angle of the arc not containing $A$ , and $\angle C$ is half the central angle of the arc not containing $C$ . The two arcs together make one full turn of $360^\circ$ , so the sum must be half of that.

The converse holds too. Any quadrilateral whose opposite angles sum to $180^\circ$ is cyclic, and that is used as a tool for showing that four points lie on one circle.

Ptolemy's theorem

The diagonals satisfy the following relation².

AC \cdot BD = AB \cdot CD + BC \cdot DA

Working it out gives the lengths below.

Length	Value
$AC$	$4\sqrt{5}$
$BD$	$7\sqrt{2}$
$AB$	$2\sqrt{5}$
$CD$	$5\sqrt{2}$
$BC$	$6$
$DA$	$3\sqrt{10}$

The left side is $4\sqrt{5} \times 7\sqrt{2} = 28\sqrt{10}$ and the right side is $10\sqrt{10} + 18\sqrt{10} = 28\sqrt{10}$ , in agreement.

For a quadrilateral that is not cyclic the equality fails, the left side being the smaller. That is Ptolemy's inequality, with equality only in the cyclic case.

Brahmagupta's formula

The area follows from the four sides alone. With $s = \dfrac{a + b + c + d}{2}$ it is as follows³.

S = \sqrt{(s - a)(s - b)(s - c)(s - d)}

For this quadrilateral $S = 42$ , agreeing with the shoelace computation $\dfrac{1}{2}(20 + 24 + 25 + 15) = 42$ . It is Heron's formula for a triangle carried over to a quadrilateral, and it applies only in the cyclic case.

The four lines on the graph are the four sides, the two arcs are the circle through the four vertices, and the large dots are the four vertices together with the center $(0, 0)$ .