## What are Inscribed Angles?

Inscribed angles are angles that start on one side of a circle and open up to subtend an arc on the opposite side of the circle. The vertex and the endpoints of the angle must be on the circumference of the circle.

## Inscribed Angle Theorem

The inscribed angle theorem says that if a central angle and an inscribed angle subtend the same arc, then the measurement of the inscribed angle will be half as big as the measurement of the central angle.

\[m\angle C=2(m\angle I)\]

## How to Use the Inscribed Angles Theorem

- Make sure the central angle and inscribed angle subtend the same arc.
- Identify which angle you know and which angle you are looking for.
- If you know the measurement of the central angle, split it in half to find the inscribed angle.
- If you know the measurement of the inscribed angle, double it to find the central angle.

## Examples

*Do \(\angle A\) and \(\angle B\) subtend the same arc?*

Yes, \(\angle A\) and \(\angle B\) have the same endpoints and they open up across the same part of the circumference.

*Were you given the measurement for the central angle or the inscribed angle?*

I know that \(m\angle A = 80^\circ\)

\(\angle A\) is the central angle and the Inscribed Angle Theorem says that the measurement of the inscribed angle is half of the measurement of the central angle.

So, that means \(m\angle B=40^\circ\)

\[m\angle B=40^\circ\]

*Do \(\angle D\) and \(\angle F\) subtend the same arc?*

Yes, \(\angle D\) and \(\angle F\) have the same endpoints and they open up across the same part of the circumference.

*Were you given the measurement for the central angle or the inscribed angle?*

I know that \(m\angle F = 50^\circ\)

\(\angle F\) is the inscribed angle and the Inscribed Angle Theorem says that the measurement of the central angle is twice the measurement of the inscribed angle.

So, that means \(m\angle D=100^\circ\)

\[m\angle D=100^\circ\]

## Diameters and Right Angles

If there is an inscribed angle that subtends the diameter of a circle, the measure of the inscribed angle will always be \(90^\circ\).

This is because the diameter is a straight line that goes through the center of the circle. This makes it a central angle of \(180^\circ\). And half of 180 is 90.

This special case of the Inscribed Angles Theorem that can be very useful when you are working with inscribed triangles.

If you know that the triangle is a right triangle, then that means you can use the Pythagorean Theorem, or sine\cosine\tangent.

## Why It Works

**Case 1**

### Case 1

### Case 2

In Case 1, we proved that the central angle is twice as big as the inscribed angle IF one of the sides of the inscribed angle was a diameter of the circle.

The Case 2 diagram does not include a diameter, but if we draw a diameter from point A through the center of the circle, we can use our results from Case 1 to prove Case 2.

When the diameter is drawn, two new angles (shown in **blue** and **red**) are created.

\({\blue c} =\) measurement of the **blue** angle

\({\red d} =\) measurement of the **red** angle

The **blue **and **red** angles subtend the **blue** arc. And the **red** angle has the diameter of the circle as one of its sides. That means that we can apply Case 1 and write this equation:

\[{\blue c}= 2{\red d}\]

If we combine the **red** and **yellow** angles into one big angle, that angle subtends the **purple** arc and it has the diameter of the circle as one of its sides.

The combination of the **blue** and **green** angles also subtends the **purple** arc.

That means we can apply Case 1 and write this equation:

\[{\blue c}+{\green a}=2({\red d}+{\yellow b})\]

\[{\blue c}= 2{\red d}\]

\[{\blue c}+{\green a}=2({\red d}+{\yellow b})\]

If I substitute the first equation into the second equation, I get…

\[2{\red d}+{\green a}=2({\red d}+{\yellow b})\]

Then I can distribute the 2 to all the terms inside the parentheses on the right side of the equation.

\[2{\red d}+{\green a}=2{\red d}+2{\yellow b}\]

Then I can subtract \(2{\red d}\) from both sides of the equation.

\[{\green a}=2{\yellow b}\]

\[{\green a}=2{\yellow b}\]

This simplified equation shows that the measure of the **central angle** is twice as big as the measure of the **inscribed angle** for all diagrams similar to this one:

### Case 3

In Case 1 of this proof, we showed that the central angle is twice as big as the inscribed angle IF one of the sides of the inscribed angle was a diameter of the circle.

The Case 3 diagram does not include a diameter, but if we draw a diameter from point A through the center of the circle, we can use our results from Case 1 to prove Case 3.

When the diameter is drawn, the **green** and **yellow** angles are split into two separate angles.

\({\green a_1} =\) measurement of the **left green** angle

\({\green a_2} =\) measurement of the **right green** angle

\({\yellow b_1} =\) measurement of the **left yellow** angle

\({\yellow b_2} =\) measurement of the **right yellow** angle

The original **green** angle is a combination of the **left green **and the **right green **angles. That means I can write this equation:

\[{\green a} = {\green a_1}+{\green a_2}\]

The original **yellow** angle is also a combination of the **left yellow**** **and the **right yellow**** **angles. That means I can write this equation:

\[{\yellow b} = {\yellow b_1}+{\yellow b_2}\]

The **left green**** **and **left yellow** angles subtend the **blue** arc. And the **left yellow** angle has the diameter of the circle as one of its sides. That means that we can apply Case 1 and write this equation:

\[{\green a_1}= 2{\yellow b_1}\]

Similarly, the **right green**** **and **right yellow** angles subtend the **purple** arc. And the **right yellow** angle has the diameter of the circle as one of its sides. That means that we can apply Case 1 and write this equation:

\[{\green a_2}= 2{\yellow b_2}\]

\[{\green a} = {\green a_1}+{\green a_2}\]

\[{\yellow b} = {\yellow b_1}+{\yellow b_2}\]

\[{\green a_1}= 2{\yellow b_1}\]

\[{\green a_2}= 2{\yellow b_2}\]

If I add the last two equations together, I get…

\[{\green a_1}+{\green a_2}= 2{\yellow b_1}+2{\yellow b_2}\]

I can factor out a 2 from both terms on the right side of the equation, which gives me…

\[{\green a_1}+{\green a_2}= 2({\yellow b_1}+{\yellow b_2})\]

Then I can substitute \({\green a}\) for \( {\green a_1}+{\green a_2}\) and \({\yellow b}\) for \({\yellow b_1}+{\yellow b_2}\).

\[{\green a}= 2({\yellow b})\]

\[{\green a}=2({\yellow b})\]

This simplified equation shows that the measure of the **central angle** is twice as big as the measure of the **inscribed angle** for all diagrams similar to this one: