Angle Theorems in Circles

Inscribed Angles Subtending the Same Arc

Inscribed angles subtending the $same arc$ have equal measures when their vertices lie on the same remaining arc, or equivalently in the same circle segment.

Inscribed Angles Subtending the Same Arc

Inscribed angles in the same segment that subtend the same arc have equal measures.

$Property$ : $\angle ACB = \angle ADB$

Both vertices $C$ and $D$ lie on the same side of chord $AB$ . The angles $\angle ACB$ and $\angle ADB$ therefore subtend the same arc from the same segment and have equal measures.

Central Angle and Inscribed Angle

The central angle is $twice the inscribed angle$ that subtends the same arc.

Relationship Between Central Angle and Inscribed Angle

Central angle

= 2 \times

inscribed angle for the same arc.

$Property$ : $\angle AOB = 2 \times \angle ACB$

Inscribed Angle Subtending a Diameter

Every inscribed angle that subtends a $diameter$ of the circle is always a right angle ( $90^\circ$ ).

Inscribed Angle Subtending a Diameter

Inscribed angle subtending a diameter is always

90^\circ

$Property$ : If $AB$ is a diameter, then $\angle ACB = \angle ADB = 90^\circ$

This is known as $Thales' Theorem$ .

Angles in a Cyclic Quadrilateral

A cyclic quadrilateral has four vertices on a circle. Two $opposite angles$ intercept arcs that together form the whole circle. Each inscribed angle is half its intercepted arc measure, so:

\alpha+\gamma=\frac{1}{2}\times360^\circ=180^\circ

Quadrilateral with Opposite Angles

Two opposite angles intercept arcs that complete the circle.

In the diagram, angle $A$ is $100^\circ$ and angle $C$ is $80^\circ$ . Neither angle has to be a right angle.

100^\circ+80^\circ=180^\circ

Exterior Angle Equals Opposite Interior Angle

Extend side $BA$ to $E$ . Exterior angle $DAE$ forms a $straight angle$ with the interior angle at $A$ . The interior angle at $C$ also adds to angle $A$ to make $180^\circ$ .

\varepsilon=180^\circ-\alpha=\gamma

Exterior and Opposite Interior Angles

Extending one side creates an exterior angle equal to the opposite interior angle.

Finding Angle Measures with the Circle Theorems

Each example applies one circle theorem to a labelled situation. Identify the $angles and arc$ connected by the theorem before calculating. Write the relation down before substituting numbers, because the relation shows which quantity is the unknown one.

Determining Angle Measures

In a circle with center $O$ , the central angle is $\angle AOB = 100^\circ$ . Find the inscribed angle $\angle ACB$ intercepting the same minor arc!

$Solution$ :

The inscribed angle intercepts the $same arc$ , so it is half the central angle:

\angle ACB = \frac{1}{2} \times \angle AOB

Finding Opposite Angles

In cyclic quadrilateral $ABCD$ , given $\angle A = 75^\circ$ . Find the measure of $\angle C$ !

$Solution$ :

The $opposite angles$ in a cyclic quadrilateral add to a straight angle:

\angle A + \angle C = 180^\circ

Using the Theorem of Thales

Point $C$ lies on a circle with $AB$ as the diameter. Find the measure of $\angle ACB$ !

$Solution$ :

Since $AB$ is a diameter and $C$ lies on the circle, by $Thales' theorem$ :

\angle ACB = 90^\circ

Practice Problems

Identify the $intercepted arc$ or the pair of opposite angles before calculating. In a diameter problem, the third vertex must differ from both diameter endpoints so that a triangle is formed.

In a circle with center $O$ , central angle $\angle AOB = 140^\circ$ intercepts the minor arc $AB$ . Points $C$ and $D$ lie on the major arc $AB$ . Find the measures of $\angle ACB$ and $\angle ADB$ .
In cyclic quadrilateral $PQRS$ , given:
$\angle P = 85^\circ$
Find the measures of $\angle R$ and $\angle S$ !
Points $A$ , $B$ , and $C$ lie on a circle. If $AB$ is a diameter and $BC = AC$ , find the measure of $\angle BAC$ !
In a circle, inscribed angle $\angle APB = 35^\circ$ . Find the measure of central angle $\angle AOB$ subtending the same arc!
In cyclic quadrilateral $KLMN$ , the exterior angle at vertex $K$ is $65^\circ$ . Find the measure of the interior angle at vertex $M$ !

Worked Solutions

$Solution 1$ : Finding inscribed angles subtending the same arc

Visualization

Central angle

\angle AOB = 140^\circ

and the inscribed angles intercept the same arc

AB

$Solution$ :

Using the relationship between central and inscribed angles:

\text{Inscribed angle} = \frac{1}{2} \times \text{Central angle}

Since inscribed angles subtending the same arc have equal measures:

\angle ADB = \angle ACB = 70^\circ

$Solution 2$ : Finding angles in a cyclic quadrilateral

$Solution$ :

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

For angle $R$ (opposite to $P$ ):

\angle P + \angle R = 180^\circ

For angle $S$ (opposite to $Q$ ):

\angle Q + \angle S = 180^\circ

$Solution 3$ : Finding angle in an isosceles triangle with diameter

Visualization

AB

is a diameter,

BC = AC

(isosceles triangle).

$Solution$ :

Since $AB$ is a diameter, by Thales' Theorem:

\angle ACB = 90^\circ

Since $BC = AC$ , triangle $ABC$ is a right isosceles triangle.

In a right isosceles triangle, both base angles are equal:

\angle BAC + \angle ABC + \angle ACB = 180^\circ

Since $\angle BAC = \angle ABC$ (base angles of isosceles triangle):

2 \times \angle BAC = 90^\circ

$Solution 4$ : Finding central angle from inscribed angle

$Solution$ :

Using the relationship between central and inscribed angles:

\text{Central angle} = 2 \times \text{Inscribed angle}

$Solution 5$ : Finding interior angle from exterior angle in cyclic quadrilateral

$Solution$ :

In a cyclic quadrilateral, an exterior angle at a vertex equals the interior angle at the opposite vertex.

If the exterior angle at $K$ is $65^\circ$ , then:

\angle M = 65^\circ

This is because vertices $K$ and $M$ are opposite in cyclic quadrilateral $KLMN$ .

Inscribed Angles Subtending the Same Arc

Inscribed angles subtending the $same arc$ have equal measures when their vertices lie on the same remaining arc, or equivalently in the same circle segment.

Inscribed Angles Subtending the Same Arc

Inscribed angles in the same segment that subtend the same arc have equal measures.

$Property$ : $\angle ACB = \angle ADB$

Both vertices $C$ and $D$ lie on the same side of chord $AB$ . The angles $\angle ACB$ and $\angle ADB$ therefore subtend the same arc from the same segment and have equal measures.

Central Angle and Inscribed Angle

The central angle is $twice the inscribed angle$ that subtends the same arc.

Relationship Between Central Angle and Inscribed Angle

Central angle

= 2 \times

inscribed angle for the same arc.

$Property$ : $\angle AOB = 2 \times \angle ACB$

Inscribed Angle Subtending a Diameter

Every inscribed angle that subtends a $diameter$ of the circle is always a right angle ( $90^\circ$ ).

Inscribed Angle Subtending a Diameter

Inscribed angle subtending a diameter is always

90^\circ

$Property$ : If $AB$ is a diameter, then $\angle ACB = \angle ADB = 90^\circ$

This is known as $Thales' Theorem$ .

Angles in a Cyclic Quadrilateral

A cyclic quadrilateral has four vertices on a circle. Two $opposite angles$ intercept arcs that together form the whole circle. Each inscribed angle is half its intercepted arc measure, so:

\alpha+\gamma=\frac{1}{2}\times360^\circ=180^\circ

Quadrilateral with Opposite Angles

Two opposite angles intercept arcs that complete the circle.

In the diagram, angle $A$ is $100^\circ$ and angle $C$ is $80^\circ$ . Neither angle has to be a right angle.

100^\circ+80^\circ=180^\circ

Exterior Angle Equals Opposite Interior Angle

Extend side $BA$ to $E$ . Exterior angle $DAE$ forms a $straight angle$ with the interior angle at $A$ . The interior angle at $C$ also adds to angle $A$ to make $180^\circ$ .

\varepsilon=180^\circ-\alpha=\gamma

Exterior and Opposite Interior Angles

Extending one side creates an exterior angle equal to the opposite interior angle.

Finding Angle Measures with the Circle Theorems

Determining Angle Measures

In a circle with center $O$ , the central angle is $\angle AOB = 100^\circ$ . Find the inscribed angle $\angle ACB$ intercepting the same minor arc!

$Solution$ :

The inscribed angle intercepts the $same arc$ , so it is half the central angle:

\angle ACB = \frac{1}{2} \times \angle AOB

Finding Opposite Angles

In cyclic quadrilateral $ABCD$ , given $\angle A = 75^\circ$ . Find the measure of $\angle C$ !

$Solution$ :

The $opposite angles$ in a cyclic quadrilateral add to a straight angle:

\angle A + \angle C = 180^\circ

Using the Theorem of Thales

Point $C$ lies on a circle with $AB$ as the diameter. Find the measure of $\angle ACB$ !

$Solution$ :

Since $AB$ is a diameter and $C$ lies on the circle, by $Thales' theorem$ :

\angle ACB = 90^\circ

Practice Problems

Identify the $intercepted arc$ or the pair of opposite angles before calculating. In a diameter problem, the third vertex must differ from both diameter endpoints so that a triangle is formed.

In a circle with center $O$ , central angle $\angle AOB = 140^\circ$ intercepts the minor arc $AB$ . Points $C$ and $D$ lie on the major arc $AB$ . Find the measures of $\angle ACB$ and $\angle ADB$ .
In cyclic quadrilateral $PQRS$ , given:
$\angle P = 85^\circ$
Find the measures of $\angle R$ and $\angle S$ !
Points $A$ , $B$ , and $C$ lie on a circle. If $AB$ is a diameter and $BC = AC$ , find the measure of $\angle BAC$ !
In a circle, inscribed angle $\angle APB = 35^\circ$ . Find the measure of central angle $\angle AOB$ subtending the same arc!
In cyclic quadrilateral $KLMN$ , the exterior angle at vertex $K$ is $65^\circ$ . Find the measure of the interior angle at vertex $M$ !

Worked Solutions

$Solution 1$ : Finding inscribed angles subtending the same arc

Visualization

Central angle

\angle AOB = 140^\circ

and the inscribed angles intercept the same arc

AB

$Solution$ :

Using the relationship between central and inscribed angles:

\text{Inscribed angle} = \frac{1}{2} \times \text{Central angle}

Since inscribed angles subtending the same arc have equal measures:

\angle ADB = \angle ACB = 70^\circ

$Solution 2$ : Finding angles in a cyclic quadrilateral

$Solution$ :

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

For angle $R$ (opposite to $P$ ):

\angle P + \angle R = 180^\circ

For angle $S$ (opposite to $Q$ ):

\angle Q + \angle S = 180^\circ

$Solution 3$ : Finding angle in an isosceles triangle with diameter

Visualization

AB

is a diameter,

BC = AC

(isosceles triangle).

$Solution$ :

Since $AB$ is a diameter, by Thales' Theorem:

\angle ACB = 90^\circ

Since $BC = AC$ , triangle $ABC$ is a right isosceles triangle.

In a right isosceles triangle, both base angles are equal:

\angle BAC + \angle ABC + \angle ACB = 180^\circ

Since $\angle BAC = \angle ABC$ (base angles of isosceles triangle):

2 \times \angle BAC = 90^\circ

$Solution 4$ : Finding central angle from inscribed angle

$Solution$ :

Using the relationship between central and inscribed angles:

\text{Central angle} = 2 \times \text{Inscribed angle}

$Solution 5$ : Finding interior angle from exterior angle in cyclic quadrilateral

$Solution$ :

In a cyclic quadrilateral, an exterior angle at a vertex equals the interior angle at the opposite vertex.

If the exterior angle at $K$ is $65^\circ$ , then:

\angle M = 65^\circ

This is because vertices $K$ and $M$ are opposite in cyclic quadrilateral $KLMN$ .