Properties of Angle in Circle

Inscribed Angles Subtending the Same Arc

Inscribed angles that subtend the same arc have equal measures, regardless of where the vertex is positioned on the circle.

Property: $\angle ACB = \angle ADB$

Both inscribed angles $\angle ACB$ and $\angle ADB$ subtend the same arc $AB$, so they 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 $C$)

Applications of Angle Properties in Circles

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 Thales' Theorem

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 $\angle AOB = 140^\circ$. If $C$ and $D$ are two different points on the circle, find:

   • The measure of angle $ACB$
   • The measure of 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 $M$!

Answer Key

1. Finding inscribed angles subtending the same arc

   Visualization

   Central angle $AOB = 140^\circ$ and inscribed angles subtending arc $AB$.

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   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, $\text{sum of opposite angles} = 180^\circ$

   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).

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   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:

   $$\angle \text{interior at M} = \angle \text{exterior at K} = 65^\circ$$

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