Angle Theorems in Circles

Inscribed Angles Subtending the Same Arc

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

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.

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 ($90^\circ$).

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 is a quadrilateral whose four vertices lie on a circle. The sum of opposite angles is $180^\circ$.

Property: $\alpha + \gamma = 180^\circ$ and $\beta + \delta = 180^\circ$

Exterior Angle Equals Opposite Interior Angle

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

Property: $\angle DAE = \angle BCD$ (exterior angle at $A$ equals interior angle at $C$)

Finding Angle Measures with the Circle Theorems

Each example applies one circle theorem to a labelled situation. Identify which angles and which arc the theorem connects 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 measure of inscribed angle $\angle ACB$!

Solution:

Using the property of central and inscribed angles:

Finding Opposite Angles

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

Solution:

Using the property of cyclic quadrilaterals:

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:

Practice Problems

1. 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 measure of angle $\angle ACB$
   • The measure of angle $\angle ADB$

2. In cyclic quadrilateral $PQRS$, given:

   $$\angle P = 85^\circ$$

   $$\angle Q = 110^\circ$$

   Find the measures of $\angle R$ and $\angle S$!

3. Points $A$, $B$, and $C$ lie on a circle. If $AB$ is a diameter and $BC = AC$, find the measure of $\angle BAC$!

4. In a circle, inscribed angle $\angle APB = 35^\circ$. Find the measure of central angle $\angle AOB$!

5. 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

1. Finding inscribed angles subtending the same arc

   Visualization

   Central angle $\angle AOB = 140^\circ$ and the inscribed angles intercept the same arc $AB$.

   GridAutomatic rotation

   Solution:

   Using the relationship between central and inscribed angles:

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

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

   $$\angle ACB = \frac{1}{2} \times 140^\circ$$

   $$\angle ACB = 70^\circ$$

   Since inscribed angles subtending the same arc have equal measures:

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

2. Finding angles in a cyclic quadrilateral

   Solution:

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

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

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

   $$85^\circ + \angle R = 180^\circ$$

   $$\angle R = 180^\circ - 85^\circ$$

   $$\angle R = 95^\circ$$

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

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

   $$110^\circ + \angle S = 180^\circ$$

   $$\angle S = 180^\circ - 110^\circ$$

   $$\angle S = 70^\circ$$

3. Finding angle in an isosceles triangle with diameter

   Visualization

   $AB$ is a diameter, $BC = AC$ (isosceles triangle).

   GridAutomatic rotation

   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:

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

   $$\angle BAC + \angle ABC + 90^\circ = 180^\circ$$

   $$\angle BAC + \angle ABC = 90^\circ$$

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

   $$2 \times \angle BAC = 90^\circ$$

   $$\angle BAC = 45^\circ$$

4. Finding central angle from inscribed angle

   Solution:

   Using the relationship between central and inscribed angles:

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

   $$\angle AOB = 2 \times \angle APB$$

   $$\angle AOB = 2 \times 35^\circ$$

   $$\angle AOB = 70^\circ$$

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 $KLMN$.