Math Core

Four points on a circle carry a lot of hidden structure: equal angles, supplementary angles, and a remarkable relation between the sides and diagonals. Spotting a cyclic quadrilateral (sometimes one that isn't drawn) is often the whole key to an AIME geometry problem.

Recognizing cyclic quadrilaterals

A quadrilateral is cyclic if all four vertices lie on one circle. Each of these conditions is equivalent to $ABCD$ (convex) being cyclic:

Opposite angles are supplementary: $\angle A + \angle C = 180^\circ$ .
Equal angles on the same side: $\angle ABD = \angle ACD$ (a side seen from the two other vertices at equal angles).
Power of a point: diagonals $AC$ and $BD$ meet at $P$ with $PA \cdot PC = PB \cdot PD$ .

All three come from the inscribed angle theorem: an inscribed angle is half the arc it cuts off. For (1), $\angle A$ and $\angle C$ cut off arcs that together make the whole circle, so they add to $\tfrac{1}{2} \cdot 360^\circ$ . The converses hold because, for a fixed segment $BD$ , the points $X$ with $\angle BXD$ equal to a given angle form an arc of a circle through $B$ and $D$ .

A classic hidden circle: if $\angle BEC = \angle BFC = 90^\circ$ , then $E$ and $F$ lie on the circle with diameter $BC$ .

Ptolemy's theorem

In a cyclic quadrilateral $ABCD$ , the product of the diagonals equals the sum of the products of opposite sides:

$AC \cdot BD = AB \cdot CD + AD \cdot BC.$

Proof. Choose $E$ on diagonal $BD$ with $\angle BAE = \angle CAD$ . Also $\angle ABE = \angle ABD = \angle ACD$ (both subtend arc $AD$ ). So $\triangle ABE \sim \triangle ACD$ (angle-angle), which gives

$\frac{AB}{AC} = \frac{BE}{CD} \quad\Longrightarrow\quad AB \cdot CD = AC \cdot BE.$

Next, $\angle BAC = \angle BAE + \angle EAC = \angle CAD + \angle EAC = \angle EAD$ , and $\angle ACB = \angle ADB = \angle ADE$ (both subtend arc $AB$ ). So $\triangle ABC \sim \triangle AED$ , which gives

$\frac{BC}{ED} = \frac{AC}{AD} \quad\Longrightarrow\quad AD \cdot BC = AC \cdot ED.$

Adding: $AB \cdot CD + AD \cdot BC = AC(BE + ED) = AC \cdot BD$ . $\blacksquare$

(For a quadrilateral that is not cyclic, the same construction gives the strict inequality $AC \cdot BD < AB \cdot CD + AD \cdot BC$ .)

A cyclic quadrilateral ABCD (A at the top, going counterclockwise) with both diagonals drawn.

Worked example: A point on the circumcircle of an equilateral triangle

Equilateral triangle $ABC$ is inscribed in a circle, and $P$ is on the arc $BC$ not containing $A$ . Show that $PA = PB + PC$ .

$ABPC$ is cyclic in that order, with diagonals $AP$ and $BC$ . Ptolemy with side $s$ :

$PA \cdot s = PB \cdot s + PC \cdot s \quad\Longrightarrow\quad PA = PB + PC.$

Two more formulas

The diagonal ratio. Write $a = AB$ , $b = BC$ , $c = CD$ , $d = DA$ and let $R$ be the circumradius. Any triangle has area $\dfrac{xyz}{4R}$ (from $\text{area} = \tfrac12 xy\sin Z$ and $z = 2R\sin Z$ ). Split the quadrilateral along each diagonal:

$[ABCD] = \frac{ab \cdot AC}{4R} + \frac{cd \cdot AC}{4R} = \frac{ad \cdot BD}{4R} + \frac{bc \cdot BD}{4R}.$

$\frac{AC}{BD} = \frac{ad + bc}{ab + cd}.$

Together with Ptolemy ( $AC \cdot BD = ac + bd$ ), this gives each diagonal from the four sides.

Brahmagupta's formula. A cyclic quadrilateral with sides $a, b, c, d$ and semiperimeter $s$ has area

$K = \sqrt{(s - a)(s - b)(s - c)(s - d)}.$

Sketch of proof: $K = \tfrac12 (ad + bc)\sin A$ because $\sin C = \sin A$ . The law of cosines on diagonal $BD$ from both sides, with $\cos C = -\cos A$ , gives $a^2 + d^2 - 2ad\cos A = b^2 + c^2 + 2bc\cos A$ . Solve for $\cos A$ , substitute into $16K^2 = 4(ad + bc)^2(1 - \cos^2 A)$ , and factor the difference of squares twice. Setting $d = 0$ recovers Heron's formula.

Worked example: Brahmagupta

A cyclic quadrilateral has sides $25$ , $39$ , $52$ , $60$ . Find its area.

$s = 88$ , so $K = \sqrt{63 \cdot 49 \cdot 36 \cdot 28} = \sqrt{3111696} = 1764$ .

There's a reason it's so clean: $25^2 + 60^2 = 39^2 + 52^2 = 65^2$ , so the diagonal between the sides $25, 60$ and $39, 52$ is a diameter of length $65$ , and the quadrilateral is two right triangles with areas $750$ and $1014$ .

Worked example: A regular heptagon identity

A regular heptagon has side $a$ , short diagonal $b$ and long diagonal $c$ . Prove that $\dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{c}$ .

Label the vertices $V_1, \dots, V_7$ and apply Ptolemy to cyclic quadrilateral $V_1V_2V_3V_5$ . Its sides are $V_1V_2 = a$ , $V_2V_3 = a$ , $V_3V_5 = b$ , $V_5V_1 = c$ , and its diagonals are $V_1V_3 = b$ and $V_2V_5 = c$ . Ptolemy:

$b \cdot c = a \cdot b + a \cdot c.$

Divide by $abc$ : $\dfrac{1}{a} = \dfrac{1}{c} + \dfrac{1}{b}$ .

Common mistake

Ptolemy pairs opposite sides, and the vertices must be taken in order around the circle. Before applying it, write the four points in circular order and identify which segments are the diagonals.

Tip

Right angles make circles. Two right angles looking at the same segment (like the feet of two altitudes) put four points on a circle with that segment as diameter. Then triangle $AEF$ (with $E$ , $F$ the feet of the altitudes from $B$ and $C$ ) is similar to triangle $ABC$ with ratio $\cos A$ .

Practice

Practice 1

A quadrilateral inscribed in a circle has sides $7$ , $15$ , $20$ and $24$ , in some order. Find its area.