# Cyclic Quadrilaterals

A **cyclic quadrilateral** is a quadrilateral that can be inscribed in a circle, meaning that there exists a circle that passes through all four vertices of the quadrilateral.

Cyclic quadrilaterals are useful in various types of geometry problems, particularly those in which angle chasing is required. It is not unusual, for instance, to intentionally add points (and lines) to diagrams in order to exploit the properties of cyclic quadrilaterals.

## Properties of Angles

The angles of cyclic quadrilaterals satisfy several important relations, as they are all inscribed angles of the circumcircle. More specifically, by the inscribed angle theorem,

\[\begin{array}{lllll} &\angle ADB = \frac{\overset{\frown}{ACB}}{2}, &\angle DBC = \frac{\overset{\frown}{CAD}}{2}, &\angle BCA = \frac{\overset{\frown}{ADB}}{2}, &\angle CAD = \frac{\overset{\frown}{DBC}}{2},\\\\ & &\angle ABC = \frac{\overset{\frown}{AC}}{2}, &\angle ABD = \frac{\overset{\frown}{AD}}{2},&&\\ & &\angle DCA = \frac{\overset{\frown}{DA}}{2}, &\angle DCB = \frac{\overset{\frown}{DB}}{2},&&\\ & &\angle BAD = \frac{\overset{\frown}{BD}}{2}, &\angle BAC = \frac{\overset{\frown}{BC}}{2},&&\\ & &\angle CDB = \frac{\overset{\frown}{CB}}{2}, &\angle CDA = \frac{\overset{\frown}{CA}}{2},&& \end{array}\]

which leads to the following two results:

**Opposite Angles**

The opposite angles of a cyclic quadrilateral add to \(180^{\circ}\), or \(\pi\) radians.

**Diagonal Angles**

In a cyclic quadrilateral \(ACBD\), we have

\[\angle ABC = \angle ADC\]

and similar relations \((\)e.g. \(\angle BCD = \angle BAD).\)

These can both be directly verified from the above angle equalities.

Also recall that \(\overset{\frown}{AB} = \angle AOB\), where \(O\) is the center of the circle, by the inscribed angle theorem. This can also lead to useful information, if the center of the circumcircle is relevant.

## Properties of Lengths

The sides and diagonals of a cyclic quadrilateral are closely related:

In cyclic quadrilateral \(ACBD\),

\[AB \cdot CD = AC \cdot BD + BC \cdot AD.\]

In other words, the product of the lengths of the diagonals is equal to the sum of the products of opposite sides.

In fact, it is true of *any* quadrilateral that
\[AB \cdot CD \leq AC \cdot BD + BC \cdot AD.\]
meaning that the cyclic quadrilateral is the equality case of this inequality.

In fact, more can be said about the diagonals: if \(a,b,c,d\) are the lengths of the sides of the quadrilateral (in clockwise order),

\[\begin{align} p&=\sqrt{\frac {(ab+cd)(ac+bd)}{ad+bc}}\\ q&=\sqrt{\frac {(ac+bd)(ad+bc)}{ab+cd}}, \end{align}\]

which also demonstrates Ptolemy's theorem.

The cyclic quadrilateral is the equality case of another inequality: given four side lengths, the cyclic quadrilateral maximizes the resulting area.

Let a cyclic quadrilateral have side lengths \(a,b,c,d\), and let \(s=\frac{a+b+c+d}{2}\) be called the semiperimeter. Then the area of the quadrilateral is equal to

\[\sqrt{(s-a)(s-b)(s-c)(s-d)}.\]

It is worth noting that in the degenerate case where one side length is zero, the above formula reduces to Heron's formula for triangles. Both of these are special cases of Bretschneider's formula.

The area is then given by a special case of Bretschneider's formula.

## See Also

**Cite as:**Cyclic Quadrilaterals.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/cyclic-quadrilaterials/