Space, shape and measurement: Explore, interpret and justify geometric relationships
Unit 3: Properties of cyclic quadrilaterals
Dylan Busa

Unit 3 outcomes
By the end of this unit you will be able to:
- Define a cyclic quadrilateral.
- Apply the theorem opposite angles of a cyclic quadrilateral are supplementary.
- Apply the theorem exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
- Apply the converses of equality between opposite angles, between angles in the same segment, and between exterior angles and interior opposite angles, to prove a quadrilateral is cyclic.
What you should know
Before you start this unit, make sure you can:
- State and use all the circle theorems covered in unit 2:
- A line drawn perpendicular to a chord from the centre of the circle bisects the chord.
- A line drawn from the circle centre to the mid-point of chord is perpendicular to the chord.
- The angle subtended by an arc or chord at the centre of a circle is twice the size of the angle subtended at the circumference.
- The diameter of a circle subtends a right angle at the circumference.
- If an angle subtended by a chord at a point on the circumference is a right angle, then the chord is a diameter.
- Angles subtended by the same arc or chord in the same segment of a circle (on the same side of the chord) are equal.
Introduction
A quadrilateral is any flat four-sided figure. Each of the four sides must be straight. A square is an example of a quadrilateral, as is a parallelogram. These are very special kinds of quadrilaterals with special properties.
Most quadrilaterals have no special characteristics other than that they have four straight sides and, therefore, four interior angles. We call these irregular quadrilaterals. Figure 1 shows various examples of quadrilaterals.

But some quadrilaterals are just the right shape that their four corners all lie on the circumference of the same circle. These are called cyclic quadrilaterals (see figure 2). Some special quadrilaterals such as squares, rectangles and parallelograms are always cyclic but many irregular quadrilaterals are cyclic as well.


Take note!
For a quadrilateral to be cyclic, all four vertices (corners) of the quadrilateral must lie on the circumference of the same circle.
Cyclic quadrilateral theorems
You do not need to be able to prove any of the cyclic quadrilateral theorems yourself. You can simply assume that they are true. The following sections explain the theorems that you need to be able to state and use. Note that they are numbered only for reference purposes and continue the numbering from unit 2. These theorems do not have official numbers.
Theorem 5
Let’s look at the theorem involving opposite angles of a cyclic quadrilateral.

Example 3.1
Given the circle with centre OO with diameter GHGHand cyclic quadrilateral FGHIFGHI. GIGI is drawn and GˆHI=66∘G^HI=66∘. Determine the values of xx, yy and zz.
Solution
x=90∘(∠s in semi-circle)x=90∘(∠s in semi-circle)
y=180∘−66∘−90∘=24∘(∠s in Δ suppl)y=180∘−66∘−90∘=24∘(∠s in Δ suppl)
z=180∘−66∘=114∘(∠s in cyclic quad)z=180∘−66∘=114∘(∠s in cyclic quad)
Theorem 6
The next theorem looks at exterior angles.

Example 3.2
Proving a quadrilateral is a cyclic quadrilateral
So far, we have seen two ways in which we can prove that a quadrilateral is a cyclic quadrilateral.
- If we can prove that the opposite interior angles of the quadrilateral are supplementary, then the quadrilateral is cyclic.
- If we can prove that the exterior angle of the quadrilateral is equal to the opposite interior angle, then the quadrilateral is cyclic.
But there is a third way. Remember theorem 4 from unit 2? It stated that if the angles subtended by a chord of the circle are on the same side of the chord, then the angles are equal.

Because ˆQ and ˆP are both subtended by arc AB (or chord AB), then we know that ˆQ=ˆP. But now have a look at ABPQ. Can you see that all four vertices lie on the circumference of the same circle? Therefore, it is a cyclic quadrilateral.

This means that we can use the converse of theorem 4 to prove that a quadrilateral is cyclic. If the angles in the same segment of a circle are equal, then the quadrilateral made by the chord and the two angles must be a cyclic quadrilateral.

Take note!

Example 3.3

Exercise 3.1
- Find the value of the unknown angles:
- In each case, determine if ABCD is a cyclic quadrilateral:
The full solutions are at the end of the unit.
Summary
In this unit you have learnt the following:
- That a cyclic quadrilateral is any four-sided shape whose vertices all lie on the circumference of the same circle.
- The opposite angles of a cyclic quadrilateral are supplementary.
- The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
- A quadrilateral can be proven to be a cyclic quadrilateral if you can show that:
- the opposite angles are supplementary
- the exterior angle is equal to the interior opposite angle
- the angles subtended by one side of the quadrilateral are equal.
Suggested time to complete: 25 minutes
- O is the centre of the circle with diameter AB. CD⊥AB at P and chord DEcuts AB at F. CB=DB.
Prove that:- CˆBP=DˆBP
- CˆED=2×CˆBA
- AˆBD=12CˆOA
- A, B, C and D are points on circle centre O. AˆBC=69∘. AD and BC are extended to meet at E. AB=AC.
- Determine, with reasons, two more angles equal to AˆBC.
- If BˆAD=82∘, calculate BˆCD, CˆED and AˆBD.
The full solutions are at the end of the unit.
Exercise 3.1
- .
- .
x=180∘−74∘(opp ∠s in cyclic quad)=106∘
y=180∘−85∘(opp ∠s in cyclic quad)=95∘ - .
ˆJ=180∘−86∘(opp ∠s in cyclic quad)=94∘
x=180∘−94∘−35∘(∠s in Δ suppl)=51∘ - .
In ΔJKM:
KˆMJ=42∘(isosc Δ)∴KˆJM=180∘−42∘−42∘(∠s in Δ suppl)=96∘x=KˆJM=96∘(ext ∠ = opp int ∠ in cyclic quad)
- .
- .
- .
CˆED=180∘−110∘(∠s on a str line)=70∘BˆDC=180∘−70∘−49∘(∠s in Δ suppl)=61∘
∴BˆAE=CˆDE
Therefore ABCD is a cyclic quadrilateral (∠s in same segment). - .
CˆBD=31∘(isosc Δ)∴BˆCD=180∘−31∘−31∘(∠s in Δ suppl)=118∘
BˆCD+BˆAD=118∘+54∘=172∘
Therefore ABCD is not a cyclic quadrilateral.
- .
Unit 3: Assessment
- .
- .
CB=DB(given)∴BˆCD=CˆDB(isosc Δ)DˆPO=180∘−90∘=90∘(∠s on str line suppl)DˆPO=CˆPOIn Δ CBP:CˆBP=180∘−CˆPO−BˆCD(∠s in Δ suppl)=90∘−BˆCDIn Δ DBP:DˆBP=180∘−DˆPO−CˆDB(∠s in Δ suppl)=90∘−CˆDB∴CˆBP=DˆBP - .
CˆED=CˆBD(∠s in same seg)But CˆBP=DˆBP(proven in a.)∴2×CˆBA=CˆBD∴CˆED=2×CˆBA - .
CˆOA=2×CˆBA(∠ at centre= 2∠ at circumference)But CˆBP=DˆBP(proven in a.)∴DˆBP=12CˆOA
- .
- .
- AˆCB=AˆBC=69∘(isosc Δ)
CˆDE=AˆBC=69∘(ext ∠= opp int ∠ in cyclic quad) - If BˆAD=82∘
BˆCD=180∘−BˆAD(opp ∠s in cyclic quad)∴BˆCD=180∘−82∘=98∘
CˆDE=69∘(proven in a.)DˆCE=BˆAD=82∘(ext ∠= opp int ∠ in cyclic quad)CˆED=180−CˆDE−DˆCE(∠s in Δ suppl)∴CˆED=180∘−69∘−82∘=29∘
AˆCB=69∘(proven in a.)DˆCE=82∘(proven above)AˆCD=180∘−AˆCB−DˆCE(∠s on a str line suppl)∴AˆCD=180∘−69∘−82∘=29∘AˆBD=AˆCD=29∘(∠s in same seg)
- AˆCB=AˆBC=69∘(isosc Δ)
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