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Cyclic Quadrilateral


A Quadrilateral for which a Circle can be circumscribed so that it touches each Vertex. The Area is then given by a special case of Bretschneider's Formula. Let the sides have lengths $a$, $b$, $c$, and $d$, let $s$ be the Semiperimeter

s\equiv {\textstyle{1\over 2}}(a+b+c+d),
\end{displaymath} (1)

and let $R$ be the Circumradius. Then
$\displaystyle A$ $\textstyle =$ $\displaystyle \sqrt{(s-a)(s-b)(s-c)(s-d)}$ (2)
  $\textstyle =$ $\displaystyle {\sqrt{(ac+bd)(ad+bc)(ab+cd)}\over 4R}\,.$ (3)

Solving for the Circumradius gives
R={\textstyle{1\over 4}}\sqrt{(ac+bd)(ad+bc)(ab+cd)\over (s-a)(s-b)(s-c)(s-d)}\,.
\end{displaymath} (4)

The Diagonals of a cyclic quadrilateral have lengths
p=\sqrt{(ab+cd)(ac+bd)\over ad+bc}
\end{displaymath} (5)

q=\sqrt{(ac+bd)(ad+bc)\over ab+cd},
\end{displaymath} (6)

so that $pq=ac+bd$. In general, there are three essentially distinct cyclic quadrilaterals (modulo Rotation and Reflection) whose edges are permutations of the lengths $a$, $b$, $c$, and $d$. Of the six corresponding Diagonal lengths, three are distinct. In addition to $p$ and $q$, there is therefore a ``third'' Diagonal which can be denoted $r$. It is given by the equation
r=\sqrt{(ad+bc)(ab+cd)\over ac+bd}.
\end{displaymath} (7)

This allows the Area formula to be written in the particularly beautiful and simple form
A={pqr\over 4R}.
\end{displaymath} (8)

The Diagonals are sometimes also denoted $p$, $q$, and $r$.

The Area of a cyclic quadrilateral is the Maximum possible for any Quadrilateral with the given side lengths. Also, the opposite Angles of a cyclic quadrilateral sum to $\pi$ Radians (Dunham 1990).

A cyclic quadrilateral with Rational sides $a$, $b$, $c$, and $d$, Diagonals $p$ and $q$, Circumradius $R$, and Area $A$ is given by $a=25$, $b=33$, $c=39$, $d=65$, $p=60$, $q=52$, $R=65/2$, and $A=1344$.


Let $AHBO$ be a Quadrilateral such that the angles $\angle HAB$ and $\angle HOB$ are Right Angles, then $AHBO$ is a cyclic quadrilateral (Dunham 1990). This is a Corollary of the theorem that, in a Right Triangle, the Midpoint of the Hypotenuse is equidistant from the three Vertices. Since $M$ is the Midpoint of both Right Triangles $\Delta AHB$ and $\Delta BOH$, it is equidistant from all four Vertices, so a Circle centered at $M$ may be drawn through them. This theorem is one of the building blocks of Heron's derivation of Heron's Formula.


Place four equal Circles so that they intersect in a point. The quadrilateral $ABCD$ is then a cyclic quadrilateral (Honsberger 1991). For a Convex cyclic quadrilateral $Q$, consider the set of Convex cyclic quadrilaterals $Q_{\vert\vert}$ whose sides are Parallel to $Q$. Then the $Q_{\vert\vert}$ of maximal Area is the one whose Diagonals are Perpendicular (Gürel 1996).

See also Bretschneider's Formula, Concyclic, Cyclic Polygon, Cyclic Quadrangle, Euler Brick, Heron's Formula, Ptolemy's Theorem, Quadrilateral


Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 123, 1987.

Dunham, W. Journey Through Genius: The Great Theorems of Mathematics. New York: Wiley, p. 121, 1990.

Gürel, E. Solution to Problem 1472. ``Maximal Area of Quadrilaterals.'' Math. Mag. 69, 149, 1996.

Honsberger, R. More Mathematical Morsels. Washington, DC: Math. Assoc. Amer., pp. 36-37, 1991.

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© 1996-9 Eric W. Weisstein