# Nakafa Learning Content

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URL: https://nakafa.com/en/subjects/mathematics/circle/properties-of-angle-in-circle
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/mathematics/circle/properties-of-angle-in-circle/en.mdx

Explore angle properties in circles including inscribed angles, Thales' theorem, and cyclic quadrilaterals. Learn theorems with clear proofs.

---

## 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.

Component: LineEquation
Props:
- title: Inscribed Angles Subtending the Same Arc
- description: All inscribed angles subtending the same arc have equal measures.
- data: [
{
points: Array.from({ length: 361 }, (_, i) => {
const angle = (i * Math.PI) / 180;
return {
x: 4 * Math.cos(angle),
y: 4 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PURPLE"),
showPoints: false,
},
{
points: [
{ x: 4, y: 0, z: 0 },
{ x: 4 * Math.cos(Math.PI / 4), y: 4 * Math.sin(Math.PI / 4), z: 0 },
],
color: getColor("ORANGE"),
showPoints: true,
labels: [
{ text: "A", at: 0, offset: [0.5, 0, 0] },
{ text: "B", at: 1, offset: [0.3, 0.3, 0] },
],
},
{
points: [
{ x: 4, y: 0, z: 0 },
{
x: 4 * Math.cos((5 * Math.PI) / 4),
y: 4 * Math.sin((5 * Math.PI) / 4),
z: 0,
},
],
color: getColor("CYAN"),
showPoints: true,
labels: [{ text: "C", at: 1, offset: [-0.5, -0.3, 0] }],
},
{
points: [
{ x: 4 * Math.cos(Math.PI / 4), y: 4 * Math.sin(Math.PI / 4), z: 0 },
{
x: 4 * Math.cos((5 * Math.PI) / 4),
y: 4 * Math.sin((5 * Math.PI) / 4),
z: 0,
},
],
color: getColor("CYAN"),
showPoints: false,
},
{
points: [
{ x: 4, y: 0, z: 0 },
{
x: 4 * Math.cos((3 * Math.PI) / 4),
y: 4 * Math.sin((3 * Math.PI) / 4),
z: 0,
},
],
color: getColor("TEAL"),
showPoints: true,
labels: [{ text: "D", at: 1, offset: [-0.5, 0.3, 0] }],
},
{
points: [
{ x: 4 * Math.cos(Math.PI / 4), y: 4 * Math.sin(Math.PI / 4), z: 0 ... [truncated; 1601 chars]
- cameraPosition: [0, 0, 12]
- showZAxis: false

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

Visible text: **Property:**

Both inscribed angles $$\angle ACB$$ and{" "}
$$\angle ADB$$ subtend the same arc $$AB$$, so they have equal measures.

Visible text: Both inscribed angles and{" "}
 subtend the same arc , so they have equal measures.

## Central Angle and Inscribed Angle

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

Component: LineEquation
Props:
- title: Relationship Between Central Angle and Inscribed Angle
- description: Central angle $$= 2 \times$$ inscribed angle for the
same arc.
  Visible text: Central angle inscribed angle for the
same arc.
- data: [
{
points: Array.from({ length: 361 }, (_, i) => {
const angle = (i * Math.PI) / 180;
return {
x: 4 * Math.cos(angle),
y: 4 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PURPLE"),
showPoints: false,
},
{
points: [
{ x: 0, y: 0, z: 0 },
{ x: 4, y: 0, z: 0 },
],
color: getColor("ORANGE"),
showPoints: true,
labels: [
{ text: "O", at: 0, offset: [-0.5, -0.5, 0] },
{ text: "A", at: 1, offset: [0.5, 0, 0] },
],
},
{
points: [
{ x: 0, y: 0, z: 0 },
{ x: 4 * Math.cos(Math.PI / 4), y: 4 * Math.sin(Math.PI / 4), z: 0 },
],
color: getColor("ORANGE"),
showPoints: true,
labels: [{ text: "B", at: 1, offset: [0.3, 0.3, 0] }],
},
{
points: [
{ x: 4, y: 0, z: 0 },
{
x: 4 * Math.cos((5 * Math.PI) / 4),
y: 4 * Math.sin((5 * Math.PI) / 4),
z: 0,
},
],
color: getColor("CYAN"),
showPoints: true,
labels: [{ text: "C", at: 1, offset: [-0.5, -0.3, 0] }],
},
{
points: [
{ x: 4 * Math.cos(Math.PI / 4), y: 4 * Math.sin(Math.PI / 4), z: 0 },
{
x: 4 * Math.cos((5 * Math.PI) / 4),
y: 4 * Math.sin((5 * Math.PI) / 4),
z: 0,
},
],
color: getColor("CYAN"),
showPoints: false,
},
{
points: Array.from({ length: 16 }, (_, i) => {
const angle = (((i * 45) / 15) * Math.PI) / 180;
return {
x: 1.2 * Math.cos(angle),
y: ... [truncated; 1992 chars]
- cameraPosition: [0, 0, 12]
- showZAxis: false

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

Visible text: **Property:**

## Inscribed Angle Subtending a Diameter

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

Visible text: Every inscribed angle that subtends a diameter of the circle is always a right angle ().

Component: LineEquation
Props:
- title: Inscribed Angle Subtending a Diameter
- description: Inscribed angle subtending a diameter is always $$90^\circ$$.
  Visible text: Inscribed angle subtending a diameter is always .
- data: [
{
points: Array.from({ length: 361 }, (_, i) => {
const angle = (i * Math.PI) / 180;
return {
x: 4 * Math.cos(angle),
y: 4 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PURPLE"),
showPoints: false,
},
{
points: [
{ x: -4, y: 0, z: 0 },
{ x: 4, y: 0, z: 0 },
],
color: getColor("ORANGE"),
showPoints: true,
labels: [
{ text: "A", at: 0, offset: [-0.5, 0, 0] },
{ text: "B", at: 1, offset: [0.5, 0, 0] },
],
},
{
points: [
{ x: -4, y: 0, z: 0 },
{ x: 0, y: 4, z: 0 },
],
color: getColor("CYAN"),
showPoints: true,
labels: [{ text: "C", at: 1, offset: [0, 0.5, 0] }],
},
{
points: [
{ x: 4, y: 0, z: 0 },
{ x: 0, y: 4, z: 0 },
],
color: getColor("CYAN"),
showPoints: false,
},
{
points: [{ x: 0, y: 0, z: 0 }],
color: getColor("AMBER"),
showPoints: true,
labels: [{ text: "O", at: 0, offset: [0, -0.5, 0] }],
},
{
points: (() => {
const C = { x: 0, y: 4 };
const A = { x: -4, y: 0 };
const B = { x: 4, y: 0 };

// Calculate angle ACB at point C (should be 90°)
const angleCA = Math.atan2(A.y - C.y, A.x - C.x);
const angleCB = Math.atan2(B.y - C.y, B.x - C.x);

return Array.from({ length: 16 }, (_, i) => {
const t = i / 15;
const angle = angleCA + t * (angleCB - angleCA);
return {
x: C.x + 0.8 * ... [truncated; 1768 chars]
- cameraPosition: [0, 0, 12]
- showZAxis: false

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

Visible text: **Property:** If is a diameter, then

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 $$180^\circ$$.

Visible text: A cyclic quadrilateral is a quadrilateral whose four vertices lie on a circle. The sum of opposite angles is .

Component: LineEquation
Props:
- title: Cyclic Quadrilateral
- description: Sum of opposite angles is $$180^\circ$$.
  Visible text: Sum of opposite angles is .
- data: [
{
points: Array.from({ length: 361 }, (_, i) => {
const angle = (i * Math.PI) / 180;
return {
x: 4 * Math.cos(angle),
y: 4 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PURPLE"),
showPoints: false,
},
{
points: [
{ x: 4, y: 0, z: 0 },
{ x: 0, y: 4, z: 0 },
{ x: -4, y: 0, z: 0 },
{ x: 0, y: -4, z: 0 },
{ x: 4, y: 0, z: 0 },
],
color: getColor("ORANGE"),
smooth: false,
showPoints: true,
labels: [
{ text: "A", at: 0, offset: [0.5, 0, 0] },
{ text: "B", at: 1, offset: [0, 0.5, 0] },
{ text: "C", at: 2, offset: [-0.5, 0, 0] },
{ text: "D", at: 3, offset: [0, -0.5, 0] },
],
},
{
points: Array.from({ length: 16 }, (_, i) => {
const angle = (((i * 90) / 15) * Math.PI) / 180;
return {
x: 0.8 * Math.cos(angle),
y: 0.8 * Math.sin(angle),
z: 0,
};
}),
color: getColor("CYAN"),
showPoints: false,
labels: [{ text: "α", at: 7, offset: [0.3, 0.2, 0] }],
},
{
points: Array.from({ length: 16 }, (_, i) => {
const angle = ((180 + (i * 90) / 15) * Math.PI) / 180;
return {
x: 0.8 * Math.cos(angle),
y: 0.8 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PINK"),
showPoints: false,
labels: [{ text: "γ", at: 7, offset: [-0.3, -0.2, 0] }],
},
]
- cameraPosition: [0, 0, 12]
- showZAxis: false

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

Visible text: **Property:** and

## Exterior Angle Equals Opposite Interior Angle

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

Component: LineEquation
Props:
- title: Exterior Angle of Cyclic Quadrilateral
- description: Exterior angle $$=$$ opposite interior angle.
  Visible text: Exterior angle opposite interior angle.
- data: [
{
points: Array.from({ length: 361 }, (_, i) => {
const angle = (i * Math.PI) / 180;
return {
x: 4 * Math.cos(angle),
y: 4 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PURPLE"),
showPoints: false,
},
{
points: [
{ x: 4, y: 0, z: 0 },
{ x: 0, y: 4, z: 0 },
{ x: -4, y: 0, z: 0 },
{ x: 0, y: -4, z: 0 },
{ x: 4, y: 0, z: 0 },
],
color: getColor("ORANGE"),
smooth: false,
showPoints: true,
labels: [
{ text: "A", at: 0, offset: [0.5, 0, 0] },
{ text: "B", at: 1, offset: [0, 0.5, 0] },
{ text: "C", at: 2, offset: [-0.5, 0, 0] },
{ text: "D", at: 3, offset: [0, -0.5, 0] },
],
},
{
points: [
{ x: 4, y: 0, z: 0 },
{ x: 6, y: 0, z: 0 },
],
color: getColor("CYAN"),
showPoints: true,
labels: [{ text: "E", at: 1, offset: [0.5, 0, 0] }],
},
{
points: Array.from({ length: 16 }, (_, i) => {
const angle = ((270 + (i * 90) / 15) * Math.PI) / 180;
return {
x: 4 + 0.8 * Math.cos(angle),
y: 0 + 0.8 * Math.sin(angle),
z: 0,
};
}),
color: getColor("PINK"),
showPoints: false,
labels: [{ text: "∠DAE", at: 7, offset: [0.8, -0.5, 0] }],
},
{
points: Array.from({ length: 16 }, (_, i) => {
const angle = ((90 + (i * 90) / 15) * Math.PI) / 180;
return {
x: -4 + 0.8 * Math.cos(angle),
y: 0 + 0.8 * Math.sin(an ... [truncated; 1328 chars]
- cameraPosition: [0, 0, 12]
- showZAxis: false

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

Visible text: **Property:** (exterior angle at equals interior angle at )

## 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 $$\angle ACB$$!

Visible text: In a circle with center , the central angle is . Find the measure of inscribed angle !

**Solution:**

Using the property of central and inscribed angles:

Component: MathContainer
Children:

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

```math
\angle ACB = \frac{1}{2} \times 100^\circ
```

```math
\angle ACB = 50^\circ
```

### Finding Opposite Angles

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

Visible text: In cyclic quadrilateral , given . Find the measure of !

**Solution:**

Using the property of cyclic quadrilaterals:

Component: MathContainer
Children:

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

```math
75^\circ + \angle C = 180^\circ
```

```math
\angle C = 180^\circ - 75^\circ
```

```math
\angle C = 105^\circ
```

### Using Thales' Theorem

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

Visible text: Point lies on a circle with AB as the diameter. Find the measure of !

**Solution:**

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

Visible text: Since is a diameter and C lies on the circle, by Thales' Theorem:

```math
\angle ACB = 90^\circ
```

## 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:

   <MathContainer>
     
   
   ```math
   \angle P = 85^\circ
   ```

     
   
   ```math
   \angle Q = 110^\circ
   ```

   </MathContainer>

   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 $$M$$!

Visible text: 1. In a circle with center , central angle . If and are two different points on the circle, find:

 - The measure of angle 
 - The measure of angle 

2. In cyclic quadrilateral , given:

 <MathContainer>
 
 

 
 

 </MathContainer>

 Find the measures of and !

3. Points , , and lie on a circle. If is a diameter and , find the measure of !

4. In a circle, inscribed angle . Find the measure of central angle !

5. In cyclic quadrilateral , the exterior angle at vertex is . Find the measure of the interior angle at vertex !

### Answer Key

1. **Finding inscribed angles subtending the same arc**

   <LineEquation
     title="Visualization"
     description={<>Central angle $$AOB = 140^\circ$$ and inscribed angles subtending arc $$AB$$.</>}
     data={[
       {
         points: Array.from({ length: 361 }, (_, i) => {
           const angle = (i * Math.PI) / 180;
           return {
             x: 4 * Math.cos(angle),
             y: 4 * Math.sin(angle),
             z: 0,
           };
         }),
         color: getColor("PURPLE"),
         showPoints: false,
       },
       {
         points: [
           { x: 0, y: 0, z: 0 },
           { x: 4, y: 0, z: 0 },
         ],
         color: getColor("ORANGE"),
         showPoints: true,
         labels: [
           { text: "O", at: 0, offset: [-0.5, -0.5, 0] },
           { text: "A", at: 1, offset: [0.5, 0, 0] },
         ],
       },
       {
         points: [
           { x: 0, y: 0, z: 0 },
           {
             x: 4 * Math.cos((140 * Math.PI) / 180),
             y: 4 * Math.sin((140 * Math.PI) / 180),
             z: 0,
           },
         ],
         color: getColor("ORANGE"),
         showPoints: true,
         labels: [{ text: "B", at: 1, offset: [-0.3, 0.5, 0] }],
       },
       {
         points: [
           { x: 4, y: 0, z: 0 },
           {
             x: 4 * Math.cos((5 * Math.PI) / 4),
             y: 4 * Math.sin((5 * Math.PI) / 4),
             z: 0,
           },
         ],
         color: getColor("CYAN"),
         showPoints: true,
         labels: [{ text: "C", at: 1, offset: [-0.5, -0.3, 0] }],
       },
       {
         points: [
           {
             x: 4 * Math.cos((140 * Math.PI) / 180),
             y: 4 * Math.sin((140 * Math.PI) / 180),
             z: 0,
           },
           {
             x: 4 * Math.cos((5 * Math.PI) / 4),
             y: 4 * Math.sin((5 * Math.PI) / 4),
             z: 0,
           },
         ],
         color: getColor("CYAN"),
         showPoints: false,
       },
       {
         points: [
           { x: 4, y: 0, z: 0 },
           {
             x: 4 * Math.cos((7 * Math.PI) / 4),
             y: 4 * Math.sin((7 * Math.PI) / 4),
             z: 0,
           },
         ],
         color: getColor("TEAL"),
         showPoints: true,
         labels: [{ text: "D", at: 1, offset: [0.3, -0.5, 0] }],
       },
       {
         points: [
           {
             x: 4 * Math.cos((140 * Math.PI) / 180),
             y: 4 * Math.sin((140 * Math.PI) / 180),
             z: 0,
           },
           {
             x: 4 * Math.cos((7 * Math.PI) / 4),
             y: 4 * Math.sin((7 * Math.PI) / 4),
             z: 0,
           },
         ],
         color: getColor("TEAL"),
         showPoints: false,
       },
       {
         points: Array.from({ length: 16 }, (_, i) => {
           const angle = (((i * 140) / 15) * Math.PI) / 180;
           return {
             x: 1.2 * Math.cos(angle),
             y: 1.2 * Math.sin(angle),
             z: 0,
           };
         }),
         color: getColor("PINK"),
         showPoints: false,
         labels: [{ text: "140°", at: 7, offset: [0.3, 0.5, 0] }],
       },
     ]}
     cameraPosition={[0, 0, 12]}
     showZAxis={false}
   />

   **Solution:**

   Using the relationship between central and inscribed angles:

   <MathContainer>
     
   
   ```math
   \text{Inscribed angle} = \frac{1}{2} \times \text{Central angle}
   ```

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

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

     
   
   ```math
   \angle ACB = 70^\circ
   ```

   </MathContainer>

   Since inscribed angles subtending the same arc have equal measures:

   
   
   ```math
   \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 $$P$$):

   <MathContainer>
     
   
   ```math
   \angle P + \angle R = 180^\circ
   ```

     
   
   ```math
   85^\circ + \angle R = 180^\circ
   ```

     
   
   ```math
   \angle R = 180^\circ - 85^\circ
   ```

     
   
   ```math
   \angle R = 95^\circ
   ```

   </MathContainer>

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

   <MathContainer>
     
   
   ```math
   \angle Q + \angle S = 180^\circ
   ```

     
   
   ```math
   110^\circ + \angle S = 180^\circ
   ```

     
   
   ```math
   \angle S = 180^\circ - 110^\circ
   ```

     
   
   ```math
   \angle S = 70^\circ
   ```

   </MathContainer>

3. **Finding angle in an isosceles triangle with diameter**

   <LineEquation
     title="Visualization"
     description={
       <>
         $$AB$$ is a diameter,{" "}
         $$BC = AC$$ (isosceles triangle).
       </>
     }
     data={[
       {
         points: Array.from({ length: 361 }, (_, i) => {
           const angle = (i * Math.PI) / 180;
           return {
             x: 4 * Math.cos(angle),
             y: 4 * Math.sin(angle),
             z: 0,
           };
         }),
         color: getColor("PURPLE"),
         showPoints: false,
       },
       {
         points: [
           { x: -4, y: 0, z: 0 },
           { x: 4, y: 0, z: 0 },
         ],
         color: getColor("ORANGE"),
         showPoints: true,
         labels: [
           { text: "A", at: 0, offset: [-0.5, 0, 0] },
           { text: "B", at: 1, offset: [0.5, 0, 0] },
         ],
       },
       {
         points: [
           { x: -4, y: 0, z: 0 },
           { x: 0, y: 4, z: 0 },
         ],
         color: getColor("CYAN"),
         showPoints: true,
         labels: [{ text: "C", at: 1, offset: [0, 0.5, 0] }],
       },
       {
         points: [
           { x: 4, y: 0, z: 0 },
           { x: 0, y: 4, z: 0 },
         ],
         color: getColor("CYAN"),
         showPoints: false,
       },
       {
         points: [{ x: 0, y: 0, z: 0 }],
         color: getColor("AMBER"),
         showPoints: true,
         labels: [{ text: "O", at: 0, offset: [0, -0.5, 0] }],
       },
       {
         points: (() => {
           const C = { x: 0, y: 4 };
           const A = { x: -4, y: 0 };
           const B = { x: 4, y: 0 };

           // Calculate angle ACB at point C (should be 90°)
           const angleCA = Math.atan2(A.y - C.y, A.x - C.x);
           const angleCB = Math.atan2(B.y - C.y, B.x - C.x);

           return Array.from({ length: 16 }, (_, i) => {
             const t = i / 15;
             const angle = angleCA + t * (angleCB - angleCA);
             return {
               x: C.x + 0.8 * Math.cos(angle),
               y: C.y + 0.8 * Math.sin(angle),
               z: 0,
             };
           });
         })(),
         color: getColor("PINK"),
         showPoints: false,
         labels: [{ text: "90°", at: 7, offset: [-0.3, -0.5, 0] }],
       },
     ]}
     cameraPosition={[0, 0, 12]}
     showZAxis={false}
   />

   **Solution:**

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

   
   
   ```math
   \angle ACB = 90^\circ
   ```

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

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

   <MathContainer>
     
   
   ```math
   \angle BAC + \angle ABC + \angle ACB = 180^\circ
   ```

     
   
   ```math
   \angle BAC + \angle ABC + 90^\circ = 180^\circ
   ```

     
   
   ```math
   \angle BAC + \angle ABC = 90^\circ
   ```

   </MathContainer>

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

   <MathContainer>
     
   
   ```math
   2 \times \angle BAC = 90^\circ
   ```

     
   
   ```math
   \angle BAC = 45^\circ
   ```

   </MathContainer>

4. **Finding central angle from inscribed angle**

   **Solution:**

   Using the relationship between central and inscribed angles:

   <MathContainer>
     
   
   ```math
   \text{Central angle} = 2 \times \text{Inscribed angle}
   ```

     
   
   ```math
   \angle AOB = 2 \times \angle APB
   ```

     
   
   ```math
   \angle AOB = 2 \times 35^\circ
   ```

     
   
   ```math
   \angle AOB = 70^\circ
   ```

   </MathContainer>

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:

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

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

Visible text: 1. **Finding inscribed angles subtending the same arc**

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 y: 4 * Math.sin((140 * Math.PI) / 180),
 z: 0,
 },
 {
 x: 4 * Math.cos((7 * Math.PI) / 4),
 y: 4 * Math.sin((7 * Math.PI) / 4),
 z: 0,
 },
 ],
 color: getColor("TEAL"),
 showPoints: false,
 },
 {
 points: Array.from({ length: 16 }, (_, i) => {
 const angle = (((i * 140) / 15) * Math.PI) / 180;
 return {
 x: 1.2 * Math.cos(angle),
 y: 1.2 * Math.sin(angle),
 z: 0,
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 />

 **Solution:**

 Using the relationship between central and inscribed angles:

 <MathContainer>
 
 

 
 

 
 

 
 

 </MathContainer>

 Since inscribed angles subtending the same arc have equal measures:

 
 

2. **Finding angles in a cyclic quadrilateral**

 **Solution:**

 In a cyclic quadrilateral, 

 For angle (opposite to ):

 <MathContainer>
 
 

 
 

 
 

 
 

 </MathContainer>

 For angle (opposite to ):

 <MathContainer>
 
 

 
 

 
 

 
 

 </MathContainer>

3. **Finding angle in an isosceles triangle with diameter**

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 title="Visualization"
 description={
 <>
 is a diameter,{" "}
 (isosceles triangle).
 </>
 }
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 {
 points: Array.from({ length: 361 }, (_, i) => {
 const angle = (i * Math.PI) / 180;
 return {
 x: 4 * Math.cos(angle),
 y: 4 * Math.sin(angle),
 z: 0,
 };
 }),
 color: getColor("PURPLE"),
 showPoints: false,
 },
 {
 points: [
 { x: -4, y: 0, z: 0 },
 { x: 4, y: 0, z: 0 },
 ],
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 showPoints: true,
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 { text: "A", at: 0, offset: [-0.5, 0, 0] },
 { text: "B", at: 1, offset: [0.5, 0, 0] },
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 { x: -4, y: 0, z: 0 },
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 { x: 4, y: 0, z: 0 },
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 points: (() => {
 const C = { x: 0, y: 4 };
 const A = { x: -4, y: 0 };
 const B = { x: 4, y: 0 };

 // Calculate angle ACB at point C (should be 90°)
 const angleCA = Math.atan2(A.y - C.y, A.x - C.x);
 const angleCB = Math.atan2(B.y - C.y, B.x - C.x);

 return Array.from({ length: 16 }, (_, i) => {
 const t = i / 15;
 const angle = angleCA + t * (angleCB - angleCA);
 return {
 x: C.x + 0.8 * Math.cos(angle),
 y: C.y + 0.8 * Math.sin(angle),
 z: 0,
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 **Solution:**

 Since is a diameter, by Thales' Theorem:

 
 

 Since , triangle is a
 right isosceles triangle.

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

 <MathContainer>
 
 

 
 

 
 

 </MathContainer>

 Since (base angles of isosceles triangle):

 <MathContainer>
 
 

 
 

 </MathContainer>

4. **Finding central angle from inscribed angle**

 **Solution:**

 Using the relationship between central and inscribed angles:

 <MathContainer>
 
 

 
 

 
 

 
 

 </MathContainer>

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 is , then:

 
 

 This is because vertices and are opposite in cyclic quadrilateral .