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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 $\left(\psi \right)$ and a central angle $\left(\theta \right)$ intercept the same arc: The measure of the central angle is double the measure of the inscribed angle.
$\theta =2\psi$

## Proof overview

To prove $\theta =2\psi$ for all $\theta$ and $\psi$ (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, $\psi$‍ .

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

Segments $\stackrel{―}{BC}$ and $\stackrel{―}{BD}$ are both radii, so they have the same length. This means that $\mathrm{△}CBD$ is isosceles, which also means that its base angles are congruent:
$m\mathrm{\angle }C=m\mathrm{\angle }D=\psi$

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

Angle $\mathrm{\angle }ABC$ is a straight angle, so
$\begin{array}{rl}\theta +m\mathrm{\angle }DBC& ={180}^{\circ }\\ \\ m\mathrm{\angle }DBC& ={180}^{\circ }-\theta \end{array}$

#### Step 3: Write an equation and solve for $\psi$‍ .

The interior angles of $\mathrm{△}CBD$ are $\psi$, $\psi$, and $\left({180}^{\circ }-\theta \right)$, and we know that the interior angles of any triangle sum to ${180}^{\circ }$.
$\begin{array}{rl}\psi +\psi +\left({180}^{\circ }-\theta \right)& ={180}^{\circ }\\ \\ 2\psi +{180}^{\circ }-\theta & ={180}^{\circ }\\ \\ 2\psi -\theta & =0\\ \\ 2\psi & =\theta \end{array}$
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, $\psi$‍ .

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

Using the diameter, let's break $\psi$ into ${\psi }_{1}$ and ${\psi }_{2}$ and $\theta$ into ${\theta }_{1}$ and ${\theta }_{2}$ 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
$\left(1\right)\phantom{\rule{1em}{0ex}}{\theta }_{1}=2{\psi }_{1}$
and
$\left(2\right)\phantom{\rule{1em}{0ex}}{\theta }_{2}=2{\psi }_{2}$
because of what we learned in Case A.

#### Step 3: Add the equations.

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: ${\theta }_{2}$ and ${\psi }_{2}$ 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:
$\left(1\right)\phantom{\rule{1em}{0ex}}{\theta }_{2}=2{\psi }_{2}$
$\left(2\right)\phantom{\rule{1em}{0ex}}\left({\theta }_{2}+\theta \right)=2\left({\psi }_{2}+\psi \right)$

#### Step 3: Substitute and simplify.

$\begin{array}{rlr}\left({\theta }_{2}+\theta \right)& =2\left({\psi }_{2}+\psi \right)& \text{(2)}\\ \\ \\ \left(2{\psi }_{2}+\theta \right)& =2\left({\psi }_{2}+\psi \right)& {\theta }_{2}=2{\psi }_{2}\\ \\ \\ 2{\psi }_{2}+\theta & =2{\psi }_{2}+2\psi \\ \\ \\ \theta & =2\psi \end{array}$
And we're done! We proved that $\theta =2\psi$ 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 $\psi$ and $\theta$. With a little algebra, we proved that $\theta =2\psi$.
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 $\theta =2\psi$.

## Want to join the conversation?

• I need help in the proofs for Case 3 in inscribed angles
• Hi Sal, I have a question about the angle theorem proof and I am curious what happened if in all cases there was a radius and the angle defined would I be able to find the arch length by using the angle proof? Or I had to identify the type of angle that I am given to figure out my arch length? Thanks....
• 5 years later... I wonder if Sal is still working on it.
• can I use ψ as a variable to measure any angle I want to?
• Yes, and it doesn't have to be an angle. You can assign any variable you like to any symbol you like. You can use Latin letters, Greek letters, Hebrew letters, random shapes, emoji, or anything else.

It's common practice to use the variables θ, φ, ψ for angle measures (I myself like to use η, since it's the letter before θ), but the rules aren't set in stone. Define whatever you like.
• What happens to the measure of the inscribed angle when its vertex is on the arc? Will it be covered in the future lecture?
• If the vertex of the inscribed angle is on the arc, then it would be the reflex of the center angle that is 2 times of the inscribed angle. You can probably prove this by slicing the circle in half through the center of the circle and the vertex of the inscribed angle then use Thales' Theorem to reach case A again (kind of a modified version of case B actually).
• is it possible to prove case c without proving a & b first?
• You do not need to prove case B to prove case C, or vice-verse. But in proving case C (or proving case B), you need to prove case A first/along the way.
• What is the greatest measure possible of an inscribed angle of a circle?
• If the angle were 180, then it would be a straight angle and the sides would form a tangent line. Anything smaller would make one side of the angle pass through a second point on the circle. So the restriction on the inscribed angle would be:
0 < ψ < 180
• Do all questions have the lines colored? If not, how would you distinguish between the two?