Math Core

Angles in circles follow one rule: every angle is measured by arcs. Once you know how inscribed angles, chord angles and tangent angles relate to the arcs they cut off, "angle chasing" becomes a matter of bookkeeping. And four points on a circle (a cyclic quadrilateral) come with extra equal angles you can use for free.

The inscribed angle theorem

An inscribed angle has its vertex on the circle and its sides along two chords. The arc between its sides (on the far side from the vertex) is the arc it subtends.

Inscribed angle theorem

An inscribed angle is half the central angle (and half the arc) it subtends:

$\angle ACB = \tfrac12 \angle AOB = \tfrac12 \overset{\frown}{AB}.$

Consequences:

Inscribed angles subtending the same arc are equal.
An angle inscribed in a semicircle is a right angle (Thales), and a right inscribed angle always stands on a diameter.

The inscribed angle at the top point C and the central angle at the center O stand on the same arc AB at the bottom. Here the arc is 110°, the central angle is 110° and the inscribed angle is 55°.

Why it works. If one side of the inscribed angle passes through the center, the triangle formed with the center is isosceles, and the exterior angle at $O$ equals twice the base angle. Every other case is a sum or difference of two such pictures.

Angles from chords, secants and tangents

Vertex	Angle equals
on the circle (inscribed, or tangent–chord)	$\tfrac12$ (intercepted arc)
inside the circle (two chords)	$\tfrac12$ (sum of the two intercepted arcs)
outside the circle (secants or tangents)	$\tfrac12$ (far arc $-$ near arc)

The tangent–chord angle counts as "on the circle": the angle between a tangent at $A$ and chord $AB$ equals half of arc $AB$ , which is the same as any inscribed angle standing on that arc.

Worked example: Inside and outside

(a) Two chords cross inside a circle. The arcs intercepted by one pair of vertical angles measure $80^\circ$ and $120^\circ$ . What are those angles? $\tfrac12(80^\circ + 120^\circ) = 100^\circ$ .

(b) Two secants from an outside point $P$ intercept a far arc of $150^\circ$ and a near arc of $50^\circ$ . Then $\angle P = \tfrac12(150^\circ - 50^\circ) = 50^\circ$ .

Cyclic quadrilaterals

A quadrilateral whose four vertices lie on one circle is cyclic.

Cyclic quadrilaterals

For a convex quadrilateral $ABCD$ , the following are equivalent:

$ABCD$ is cyclic.
Opposite angles are supplementary: $\angle A + \angle C = 180^\circ$ .
A side subtends equal angles from the other two vertices: $\angle ABD = \angle ACD$ .

So you can use these angle facts when you know a quadrilateral is cyclic, and you can prove it is cyclic by finding one of them.

Opposite angles are supplementary because $\angle A$ and $\angle C$ stand on the two arcs $BCD$ and $BAD$ , which together make the whole $360^\circ$ .

Worked example: Opposite angles

In cyclic quadrilateral $ABCD$ , $\angle A = 3x$ and $\angle C = 2x$ . Then $5x = 180^\circ$ , so $x = 36^\circ$ , $\angle A = 108^\circ$ and $\angle C = 72^\circ$ .

Worked example: A hidden cyclic quadrilateral

In acute triangle $ABC$ , altitudes $\overline{BE}$ and $\overline{CF}$ meet at $H$ . If $\angle A = 64^\circ$ , find $\angle BHC$ .

In quadrilateral $AFHE$ , the angles at $F$ and $E$ are right angles, so they add to $180^\circ$ and $AFHE$ is cyclic. Opposite angles $A$ and $FHE$ are supplementary: $\angle FHE = 116^\circ$ . Then $\angle BHC = \angle FHE = 116^\circ$ (vertical angles).

Ptolemy's Theorem

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

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

Worked example: Ptolemy in an equilateral triangle

Equilateral triangle $ABC$ is inscribed in a circle, and $P$ is on the arc $BC$ that does not contain $A$ . If $PB = 3$ and $PC = 5$ , find $PA$ .

$ABPC$ is cyclic in that order, with diagonals $AP$ and $BC$ . Let $s$ be the side. Ptolemy gives $PA \cdot s = PB \cdot s + PC \cdot s$ , so $PA = PB + PC = 8$ .

Common mistake

Use Ptolemy with the vertices in their order around the circle. The diagonals connect opposite vertices in that order. Getting the order wrong gives an equation that is simply false.

Practice

Practice 1

Triangle $ABC$ is inscribed in a circle with center $O$ , and $\angle BAC = 50^\circ$ . What is $\angle OBC$ , in degrees?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

The angles $A$ , $B$ and $C$ of cyclic quadrilateral $ABCD$ are in the ratio $2 : 3 : 4$ . What is the largest angle of the quadrilateral?

Practice 3

Two chords of a circle cross, forming a $70^\circ$ angle. The arc intercepted by this angle measures $110^\circ$ . What is the measure, in degrees, of the arc intercepted by the vertical angle opposite it?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Tangents from a point $P$ touch a circle at $A$ and $B$ , and $\angle APB = 50^\circ$ . Point $C$ lies on the major arc $AB$ . What is $\angle ACB$ ?

Practice 5

$\overline{AB}$ is a diameter of a circle of radius $13$ , and $C$ is a point on the circle with $AC = 10$ . What is the area of $\triangle ABC$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Two secants are drawn to a circle from an outside point $P$ . The far arc between them is three times the near arc, and $\angle P = 40^\circ$ . What is the measure of the far arc?

Practice 7

A regular pentagon has side length $1$ . What is the length of each of its diagonals?

Practice 8

Quadrilateral $ABCD$ is inscribed in a circle, with $AD = 3$ , $DC = 8$ and $\angle ADC = 60^\circ$ . Point $B$ is on the arc $AC$ not containing $D$ , with $BA = BC$ . What is $BD$ ?

Real contest practice

2022 AMC 10A, Problem 15: a quadrilateral inscribed in a circle; supplementary opposite angles and Thales reveal the diameter.
2015 AMC 12B, Problem 19: four points built from squares on a right triangle lie on a circle.