Circle Theorems and Geometric Proofs for ESAT Mathematics

Understanding Subtended Angles

In circle geometry, the term subtended is used to describe an angle formed when two rays pass through the endpoints of an arc, line segment, or chord. An angle can be subtended at the centre of the circle or at any point on the circumference.

Angle at the Centre and Circumference

The fundamental circle theorem states that the angle subtended at the centre of a circle by a chord is exactly twice the size of the angle subtended at the circumference by the same chord.

When applying this theorem, the direction of the angles is important. As shown in the diagrams, the theorem applies to the angle at the centre and the angle at the circumference based on the same chord $AB$. If the chord subtends a reflex angle at the centre, the angle at the circumference is still half of that reflex angle.

Worked Example: Finding the Angle Subtended at the Circumference

Points $A$, $B$, and $C$ lie on the circumference of a circle with centre $O$. The obtuse angle $AOB$ is $130^\circ$. What is the size of the marked angle $ACB$?

To solve this, we must identify the correct angle at the centre. The angle at the centre that is twice angle $ACB$ is the reflex angle $AOB$, not the obtuse $130^\circ$ angle. First, calculate the reflex angle: $360^\circ - 130^\circ = 230^\circ$. Since the angle at the centre is twice the angle at the circumference, angle $ACB = 230^\circ \div 2 = 115^\circ$.

Angle in a Semicircle

If a chord is also the diameter of the circle, it passes through the centre, making the angle at the centre $180^\circ$. Consequently, any angle subtended at the circumference by the diameter is $90^\circ$. This is often referred to as the angle in a semicircle theorem.

Worked Example: Angle in a Semicircle

In triangle $ABC$ inscribed in a circle with centre $O$, $AC$ is the diameter. If angle $BAC = 27^\circ$, find angle $BCA$.

Because $AC$ is the diameter, angle $ABC$ is the angle in a semicircle and must be $90^\circ$. Using the fact that the sum of angles in a triangle is $180^\circ$, we calculate angle $BCA = 180^\circ - 90^\circ - 27^\circ = 63^\circ$.

Angles in the Same Segment

Angles subtended at the circumference by the same chord or arc are equal, provided they are in the same segment. This is known as the angles in the same segment theorem.

If the angles are in different segments, they are not equal. As shown below, angles $x$ and $y$ are not equal because they are on opposite sides of the chord $AB$.

Worked Example: Using Multiple Theorems

Given angle $ABE = 67^\circ$ in the circle below, which other angle must be $67^\circ$, and what is the size of angle $AOE$?

Angle $ABE$ is in the segment defined by chord $AE$. Another angle subtended at the circumference by the same chord is angle $ADE$, so $ADE = 67^\circ$. Note that angle $ACE$ is not equal to $67^\circ$ because its vertex $C$ is not on the circumference. Angle $AOE$ is at the centre subtended by $AE$, so it is twice angle $ABE$. Therefore, angle $AOE = 2 \times 67^\circ = 134^\circ$.

Alternate Segment Theorem

The alternate segment theorem states that the angle between a tangent and a chord through the point of contact is equal to the angle subtended by that chord in the alternate segment.

Worked Example: Alternate Segment Theorem

$BDE$ is a triangle inscribed in a circle with tangent $ABC$ touching at $B$. Angle $BDE = 68^\circ$ and angle $EBC = 58^\circ$. Find angle $BED$.

Method 1: Since $ABC$ is a straight line, angle $ABD = 180^\circ - 68^\circ - 58^\circ = 54^\circ$. According to the alternate segment theorem, angle $ABD$ is between the tangent and chord $BD$, so it equals the angle in the alternate segment, which is angle $BED$. Thus, angle $BED = 54^\circ$.

Method 2: Angle $CBE$ is between the tangent and chord $BE$. The alternate segment angle is $BDE = 58^\circ$. In triangle $BDE$, angle $BED = 180^\circ - (68^\circ + 58^\circ) = 54^\circ$.

Radius and Tangent

The angle between a radius and a tangent at the point of contact is always $90^\circ$.

Worked Example: Radius and Tangent

$ABC$ and $ADE$ are tangents at points $B$ and $D$ to a circle with centre $O$. Angle $BAD = 82^\circ$. What is the size of angle $BOD$?

Angles $OBA$ and $ODA$ are both formed by a radius meeting a tangent, so they are both $90^\circ$. $OBAD$ is a quadrilateral, and its angles must sum to $360^\circ$. Therefore, angle $BOD = 360^\circ - (90^\circ + 90^\circ + 82^\circ) = 98^\circ$.

Properties of Cyclic Quadrilaterals

A cyclic quadrilateral is a four sided shape where all four vertices lie on the circumference of a circle. The following properties apply:

1. Opposite interior angles sum to $180^\circ$.
2. The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Worked Example: Cyclic Quadrilateral

$ABCD$ is a cyclic quadrilateral and $BAE$ is a straight line. $BC$ is parallel to $AD$. If the exterior angle $BCD$ is $74^\circ$, find angle $CBA$.

Method 1: The exterior angle of a cyclic quadrilateral equals the interior opposite angle, so angle $DAE = angle BCD = 74^\circ$. Since $BC$ is parallel to $AD$, angles $CBA$ and $DAE$ are corresponding, making angle $CBA = 74^\circ$.

Method 2: Because $BC$ and $AD$ are parallel, co-interior angles sum to $180^\circ$. Thus, angle $CDA = 180^\circ - 74^\circ = 106^\circ$. In a cyclic quadrilateral, opposite angles sum to $180^\circ$, so angle $CBA = 180^\circ - 106^\circ = 74^\circ$.

Combining Circle Theorems

Exam questions often require using multiple theorems simultaneously. For example, to find angle $BOD$ when given angle $DBC = 58^\circ$ between a tangent $ABC$ and chord $BD$:

First, identify the angle in the alternate segment. If we pick any point $E$ on the major arc, angle $BED = 58^\circ$ (Alternate Segment Theorem). Then, angle $BOD = 2 \times 58^\circ = 116^\circ$ (Angle at centre is twice angle at circumference).