Cyclic Quadrilaterals

A cyclic quadrilateral has all four vertices on a circle. Its angle properties appear in almost every Paper 2 geometry rider, and proving that a quadrilateral is cyclic is a common final step worth several marks.

Opposite angles are supplementary

Theorem. The opposite angles of a cyclic quadrilateral are supplementary. (Reason: opp $∠$ s of cyclic quad)

Proof (examinable). Let $A B C D$ be cyclic with centre $O$ . Join $O B$ and $O D$ . Let $\hat{O}_{1}$ be the angle at the centre subtended by arc $B C D$ and $\hat{O}_{2}$ the reflex angle subtended by arc $B A D$ .

$\hat{O}_{1} = 2 \hat{A}$ ( $∠$ at centre $= 2 \times ∠$ at circumference)
$\hat{O}_{2} = 2 \hat{C}$ ( $∠$ at centre $= 2 \times ∠$ at circumference)
$\hat{O}_{1} + \hat{O}_{2} = 360°$ ( $∠$ s round a pt)

So $2 \hat{A} + 2 \hat{C} = 360°$ and $\hat{A} + \hat{C} = 180°$ .

Example. If $\hat{A} = 3 x + 10°$ and $\hat{C} = 2 x - 5°$ , then $5 x + 5° = 180°$ , so $x = 35°$ .

Exterior angle equals interior opposite angle

Theorem. The exterior angle of a cyclic quadrilateral equals the interior opposite angle. (Reason: ext $∠$ of cyclic quad)

This follows from the previous theorem and angles on a straight line.

Example. $A B C D$ is cyclic and $A D$ is produced to $E$ . If $\hat{C D E} = 72°$ , then $\hat{A B C} = 72°$ .

Look for this theorem whenever a side of a cyclic quadrilateral is extended.

Proving a quadrilateral is cyclic

A quadrilateral is cyclic if any one of these is true:

A pair of opposite angles is supplementary. (Reason: converse opp $∠$ s of cyclic quad)
An exterior angle equals the interior opposite angle. (Reason: converse ext $∠$ of cyclic quad)
A side subtends equal angles at the other two vertices. (Reason: converse $∠$ s in the same seg)

Example. In $△ A B C$ , altitudes $A D$ and $B E$ meet at $H$ . In quadrilateral $C D H E$ , $\hat{D} = \hat{E} = 90°$ , so $\hat{D} + \hat{E} = 180°$ and $C D H E$ is cyclic. In $A B D E$ , $A B$ subtends $90°$ at both $D$ and $E$ , so $A B D E$ is cyclic.

State which converse you use. Writing 'cyclic quad' as the reason for a quadrilateral being cyclic is circular and scores 0.

Key terms

Cyclic quadrilateral: A quadrilateral whose four vertices lie on one circle.
Supplementary: Adding up to 180°.
Converse: A theorem with the given and conclusion swapped, used to prove a quadrilateral is cyclic.
Exterior angle: The angle between one side and the extension of the adjacent side.

Common mistakes

Using cyclic quadrilateral properties before proving the quadrilateral is cyclic.
Pairing adjacent angles instead of opposite angles.
Using the exterior angle with the wrong interior angle; it must be the opposite one.
Giving 'cyclic quad' as a reason for proving it is cyclic.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

$A B C D$ is a cyclic quadrilateral with $\hat{A} = 3 x + 10°$ and $\hat{C} = 2 x - 5°$ . Calculate $x$ and the size of $\hat{A}$ .

Worked solution

Hidden - try the question first

Solution

$3 x + 10° + 2 x - 5° = 180°$ (opp $∠$ s of cyclic quad)
$5 x = 175°$ , so $x = 35°$
$\hat{A} = 115°$

Question 2

5 marks total

knowledge

Prove the theorem that states that the opposite angles of a cyclic quadrilateral are supplementary.

Worked solution

Hidden - try the question first

Solution

Construction: join $B$ and $D$ to the centre $O$
$\hat{O}_{1} = 2 \hat{A}$ ( $∠$ at centre $= 2 \times ∠$ at circumference)
$\hat{O}_{2} = 2 \hat{C}$ ( $∠$ at centre $= 2 \times ∠$ at circumference)
$\hat{O}_{1} + \hat{O}_{2} = 360°$ ( $∠$ s round a pt)
$2 \hat{A} + 2 \hat{C} = 360°$ , so $\hat{A} + \hat{C} = 180°$

Question 3

6 marks total

complex

$A B C D$ is a cyclic quadrilateral. $A D$ is produced to $E$ . Diagonal $B D$ is drawn. $\hat{C D E} = 72°$ and $\hat{A B D} = 40°$ .

Calculate, with reasons:

(a) $\hat{A B C}$ (2)

(b) $\hat{D B C}$ (2)

Worked solution

Hidden - try the question first

Solution

(a) $B$ (ext $∠$ of cyclic quad)
$= 72°$
(b) $\hat{D B C} = \hat{A B C} - \hat{A B D}$
$= 72° - 40° = 32°$
(c) $\hat{D A C} = \hat{D B C}$ ( $∠$ s in the same seg)
$= 32°$

Question 4

9 marks total

problem-solving

In acute $△ A B C$ , $A D ⊥ B C$ with $D$ on $B C$ , and $B E ⊥ A C$ with $E$ on $A C$ . $A D$ and $B E$ meet at $H$ .

Prove that:

(a) $C D H E$ is a cyclic quadrilateral. (3)

(b) $A B D E$ is a cyclic quadrilateral. (3)

Worked solution

Hidden - try the question first

Solution

(a) $B$ (given $A D ⊥ B C$ ) and $\hat{H E C} = 90°$ (given $B E ⊥ A C$ )
$\hat{H D C} + \hat{H E C} = 180°$
$C D H E$ is cyclic (converse opp $∠$ s of cyclic quad)
(b) $\hat{A D B} = 90°$ and $\hat{A E B} = 90°$
$A B$ subtends equal angles at $D$ and $E$
$A B D E$ is cyclic (converse $∠$ s in the same seg)
(c) In cyclic quad $A B D E$ , $\hat{E D C}$ is the exterior angle at $D$
$\hat{E D C} = \hat{B A E}$ (ext $∠$ of cyclic quad)
$\hat{B A E} = \hat{B A C}$ , so $\hat{E D C} = \hat{B A C}$

FAQs

How do I prove that a quadrilateral is cyclic?

Show opposite angles are supplementary, an exterior angle equals the interior opposite angle, or a side subtends equal angles at the other two vertices. Name the converse you use.

Is the cyclic quadrilateral proof examinable?

Yes. The proof that opposite angles of a cyclic quadrilateral are supplementary is one of the examinable theorem proofs.

Can a square or rectangle be cyclic?

Yes. Every rectangle, including a square, is cyclic because its opposite angles add up to 180°.

Opposite angles are supplementary

Exterior angle equals interior opposite angle

Proving a quadrilateral is cyclic

Key terms

Common mistakes

Practice questions

FAQs

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