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## High school geometry

### Course: High school geometry>Unit 8

Lesson 7: Inscribed angles

# Inscribed angle theorem proof

Proving that an inscribed angle is half of a central angle that subtends the same arc.

## Getting started

Before we get to talking about the proof, let's make sure we understand a few fancy terms related to circles.
Here's a short matching activity to see if you can figure out the terms yourself:
Using the image, match the variables to the terms.

Nice work! We'll be using these terms through the rest of the article.

## What we're about to prove

We're about to prove that something cool happens when an inscribed angle left parenthesis, start color #11accd, \psi, end color #11accd, right parenthesis and a central angle left parenthesis, start color #aa87ff, theta, end color #aa87ff, right parenthesis intercept the same arc: The measure of the central angle is double the measure of the inscribed angle.
start color #aa87ff, theta, end color #aa87ff, equals, 2, start color #11accd, \psi, end color #11accd

## Proof overview

To prove start color #aa87ff, theta, end color #aa87ff, equals, 2, start color #11accd, \psi, end color #11accd for all start color #aa87ff, theta, end color #aa87ff and start color #11accd, \psi, end color #11accd (as we defined them above), we must consider three separate cases:
Case ACase BCase C
Together, these cases account for all possible situations where an inscribed angle and a central angle intercept the same arc.

## Case A: The diameter lies along one ray of the inscribed angle, $\blueD \psi$start color #11accd, \psi, end color #11accd.

#### Step 1: Spot the isosceles triangle.

Segments start overline, start color #e84d39, B, C, end color #e84d39, end overline and start overline, start color #e84d39, B, D, end color #e84d39, end overline are both radii, so they have the same length. This means that triangle, C, B, D is isosceles, which also means that its base angles are congruent:
m, angle, C, equals, m, angle, D, equals, start color #11accd, \psi, end color #11accd

#### Step 2: Spot the straight angle.

Angle angle, start color #e84d39, A, B, C, end color #e84d39 is a straight angle, so
\begin{aligned} \purpleC \theta + m\angle DBC &= 180^\circ \\\\ m\angle DBC &= 180^\circ - \purpleC \theta \end{aligned}

#### Step 3: Write an equation and solve for $\blueD \psi$start color #11accd, \psi, end color #11accd.

The interior angles of triangle, C, B, D are start color #11accd, \psi, end color #11accd, start color #11accd, \psi, end color #11accd, and left parenthesis, 180, degrees, minus, start color #aa87ff, theta, end color #aa87ff, right parenthesis, and we know that the interior angles of any triangle sum to 180, degrees.
\begin{aligned} \blueD{\psi} + \blueD{\psi} + (180^\circ- \purpleC{\theta}) &= 180^\circ \\\\ 2\blueD{\psi} + 180^\circ- \purpleC{\theta} &= 180^\circ \\\\ 2\blueD{\psi}- \purpleC{\theta} &=0 \\\\ 2\blueD{\psi} &=\purpleC{\theta} \end{aligned}
Cool. We've completed our proof for Case A. Just two more cases left!

## Case B: The diameter is between the rays of the inscribed angle, $\blueD \psi$start color #11accd, \psi, end color #11accd.

#### Step 1: Get clever and draw the diameter

Using the diameter, let's break start color #11accd, \psi, end color #11accd into start color #11accd, \psi, start subscript, 1, end subscript, end color #11accd and start color #11accd, \psi, start subscript, 2, end subscript, end color #11accd and start color #aa87ff, theta, end color #aa87ff into start color #aa87ff, theta, start subscript, 1, end subscript, end color #aa87ff and start color #aa87ff, theta, start subscript, 2, end subscript, end color #aa87ff as follows:

#### Step 2: Use what we learned from Case A to establish two equations.

In our new diagram, the diameter splits the circle into two halves. Each half has an inscribed angle with a ray on the diameter. This is the same situation as Case A, so we know that
left parenthesis, 1, right parenthesis, start color #aa87ff, theta, start subscript, 1, end subscript, end color #aa87ff, equals, 2, start color #11accd, \psi, start subscript, 1, end subscript, end color #11accd
and
left parenthesis, 2, right parenthesis, start color #aa87ff, theta, start subscript, 2, end subscript, end color #aa87ff, equals, 2, start color #11accd, \psi, start subscript, 2, end subscript, end color #11accd
because of what we learned in Case A.

#### Step 3: Add the equations.

\begin{aligned} \purpleC{\theta_1} + \purpleC{\theta_2} &= 2\blueD{\psi_1}+2\blueD{\psi_2}&\small \text{Add (1) and (2)} \\\\\\ (\purpleC{\theta_1} + \purpleC{\theta_2}) &= 2(\blueD{\psi_1}+\blueD{\psi_2}) &\small \text{Group variables} \\\\\\ \purpleC{\theta} &= 2\blueD{\psi} &\small\purpleC{\theta=\theta_1+\theta_2} \text{ and } \blueD{\psi=\psi_1+\psi_2} \end{aligned}
Case B is complete. Just one case left!

## Case C: The diameter is outside the rays of the inscribed angle.

#### Step 1: Get clever and draw the diameter

Using the diameter, let's create two new angles: start color #ed5fa6, theta, start subscript, 2, end subscript, end color #ed5fa6 and start color #e07d10, \psi, start subscript, 2, end subscript, end color #e07d10 as follows:

#### Step 2: Use what we learned from Case A to establish two equations.

Similar to what we did in Case B, we've created a diagram that allows us to make use of what we learned in Case A. From this diagram, we know the following:
left parenthesis, 1, right parenthesis, start color #ed5fa6, theta, start subscript, 2, end subscript, end color #ed5fa6, equals, 2, start color #e07d10, \psi, start subscript, 2, end subscript, end color #e07d10
left parenthesis, 2, right parenthesis, left parenthesis, start color #ed5fa6, theta, start subscript, 2, end subscript, end color #ed5fa6, plus, start color #aa87ff, theta, end color #aa87ff, right parenthesis, equals, 2, left parenthesis, start color #e07d10, \psi, start subscript, 2, end subscript, end color #e07d10, plus, start color #11accd, \psi, end color #11accd, right parenthesis

#### Step 3: Substitute and simplify.

\begin{aligned} (\maroonC{\theta_2} + \purpleC{\theta}) &= 2(\goldD{\psi_2} + \blueD{\psi})&\small \text{(2)} \\\\\\ (2\goldD{\psi_2} + \purpleC{\theta})&= 2(\goldD{\psi_2} + \blueD{\psi}) &\small \maroonC{\theta_2}=2\goldD{\psi_2} \\\\\\ 2\goldD{\psi_2}+ \purpleC{\theta} &= 2\goldD{\psi_2} + 2\blueD{\psi} \\\\\\ \purpleC{\theta} &= 2\blueD{\psi} \end{aligned}
And we're done! We proved that start color #aa87ff, theta, end color #aa87ff, equals, 2, start color #11accd, \psi, end color #11accd in all three cases.

## A summary of what we did

We set out to prove that the measure of a central angle is double the measure of an inscribed angle when both angles intercept the same arc.
We began the proof by establishing three cases. Together, these cases accounted for all possible situations where an inscribed angle and a central angle intercept the same arc.
Case ACase BCase C
In Case A, we spotted an isosceles triangle and a straight angle. From this, we set up some equations using start color #11accd, \psi, end color #11accd and start color #7854ab, theta, end color #7854ab. With a little algebra, we proved that start color #aa87ff, theta, end color #aa87ff, equals, 2, start color #11accd, \psi, end color #11accd.
In cases B and C, we cleverly introduced the diameter:
Case BCase C
This made it possible to use our result from Case A, which we did. In both Case B and Case C, we wrote equations relating the variables in the figures, which was only possible because of what we'd learned in Case A. After we had our equations set up, we did some algebra to show that start color #aa87ff, theta, end color #aa87ff, equals, 2, start color #11accd, \psi, end color #11accd.