Cyclic Quadrilaterals

A cyclic quadrilateral is a quadrilateral whose four vertices all lie on a single circle. Such a figure has a special angle property: opposite interior angles sum to $180^\circ$ . Equally beautiful is the converse: any quadrilateral whose opposite angles sum to $180^\circ$ is cyclic. This pair of theorems closes the chapter.

Definition

A cyclic quadrilateral is a quadrilateral whose four vertices lie on a common circle. The circle is the circumscribed circle (or simply, the circumcircle) of the quadrilateral.

Every triangle has a circumscribed circle (the perpendicular bisectors of two sides meet at the centre). But not every quadrilateral is cyclic , only those satisfying the opposite-angle condition.

The theorem and its converse

Theorem. In a cyclic quadrilateral, opposite angles sum to $180^\circ$ .

In symbols: in cyclic quadrilateral $ABCD$ , $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$ .

Converse. If the opposite angles of a quadrilateral sum to $180^\circ$ , then the quadrilateral is cyclic.

Proof of the theorem

Given. Cyclic quadrilateral $ABCD$ inscribed in a circle with centre $O$ .

To prove. $\angle A + \angle C = 180^\circ$ .

Proof. $\angle A$ is an inscribed angle subtending arc $BCD$ (the arc not containing $A$ ). $\angle C$ is an inscribed angle subtending arc $BAD$ (the arc not containing $C$ ).

By the angle-at-centre theorem, $\angle A = \tfrac{1}{2} (\text{central angle of arc } BCD), \qquad \angle C = \tfrac{1}{2} (\text{central angle of arc } BAD).$

The two arcs together make up the entire circle, so their central angles sum to $360^\circ$ . Therefore $\angle A + \angle C = \tfrac{1}{2} \cdot 360^\circ = 180^\circ.$

Q.E.D.

The proof is one line of arithmetic once the central-angle theorem is in hand.

Proof of the converse

The converse is more subtle. Sketch: suppose $\angle A + \angle C = 180^\circ$ . Draw the circle through $A, B, D$ (three non-collinear points determine a unique circle). Let $C'$ be the point where line $BC$ (or some chosen path) meets the circle. The angle $\angle BC'D$ at $C'$ on the circle satisfies $\angle A + \angle BC'D = 180^\circ$ by Theorem 1. But $\angle A + \angle C = 180^\circ$ also, so $\angle C = \angle BC'D$ . This forces $C = C'$ , i.e. $C$ is on the circle. Thus $ABCD$ is cyclic.

Three useful applications

Application 1: Constructing a circumcircle. If a quadrilateral's opposite angles sum to $180^\circ$ , you can construct a single circle through all four vertices. Otherwise, no such circle exists.

Application 2: A rectangle is cyclic. All four angles are $90^\circ$ . Opposite angles sum to $180^\circ$ . So every rectangle is cyclic. The same holds for any square.

Application 3: An isosceles trapezium is cyclic. This is a slightly subtle fact: in an isosceles trapezium, the base angles are equal, and the angle conditions can be verified to give opposite angles summing to $180^\circ$ .

Worked examples

Example 1. In cyclic quadrilateral $ABCD$ , $\angle A = 70^\circ$ . Find $\angle C$ .

$\angle C = 180 - 70 = 110^\circ$ .

Example 2. In cyclic quadrilateral $ABCD$ , $\angle A = 80^\circ$ and $\angle B = 110^\circ$ . Find $\angle C$ and $\angle D$ .

$\angle C = 180 - 80 = 100^\circ$ . $\angle D = 180 - 110 = 70^\circ$ . Verify: $80 + 110 + 100 + 70 = 360^\circ$ . ✓

Example 3. Is a parallelogram cyclic? When?

In a parallelogram, opposite angles are equal: $\angle A = \angle C$ . For the parallelogram to be cyclic, $\angle A + \angle C = 180^\circ$ also, so $2\angle A = 180^\circ$ , hence $\angle A = 90^\circ$ . The parallelogram has a right angle , making it a rectangle. So a parallelogram is cyclic iff it is a rectangle.

Example 4. In cyclic quadrilateral $ABCD$ , $\angle A : \angle C = 2 : 3$ . Find them.

$\angle A + \angle C = 180^\circ$ . Let $\angle A = 2k, \angle C = 3k$ . Then $5k = 180 \Rightarrow k = 36$ . So $\angle A = 72^\circ$ and $\angle C = 108^\circ$ .

Example 5. A rectangle is cyclic. Identify the centre of its circumscribed circle.

The centre is the intersection of the diagonals , equidistant from all four vertices (each distance equals half the diagonal length).

Try it yourself

State the cyclic-quadrilateral theorem and its converse.
In cyclic quadrilateral $ABCD$ , $\angle A = 95^\circ$ . Find $\angle C$ .
Prove that the sum of opposite angles of a cyclic quadrilateral is $180^\circ$ .
In cyclic quadrilateral $ABCD$ , $\angle B + \angle D = 180^\circ$ is true by the theorem. Why?
Is every rectangle cyclic? Why?
Is every parallelogram cyclic? Why or why not?
A quadrilateral has angles $70^\circ, 110^\circ, 80^\circ, 100^\circ$ . Is it cyclic?
A cyclic quadrilateral has angles in ratio $1 : 2 : 3 : 4$ . Find them.
State the converse of the cyclic-quadrilateral theorem.
The exterior angle of a cyclic quadrilateral at any vertex equals which other angle? (Hint: linear pair + opposite-angle property.)

Pitfalls / Insight

Both pairs of opposite angles , the theorem is about each pair separately, both summing to $180^\circ$ .
Cyclic = vertices on a common circle. Not every quadrilateral is cyclic.
The converse needs care. Knowing one pair sums to $180^\circ$ is enough; if so, the other pair also sums to $180^\circ$ (since all four angles sum to $360^\circ$ ).

Insight. The cyclic-quadrilateral theorem unifies all the angle properties of inscribed figures. It is the natural endpoint of the chapter: a single condition that captures "the four points lie on a circle". Combined with the central-angle theorem, it lets you prove and recognise cyclic configurations effortlessly.