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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.
1

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 (ψ) and a central angle (θ) intercept the same arc: The measure of the central angle is double the measure of the inscribed angle.
θ=2ψ

Proof overview

To prove θ=2ψ for all θ and ψ (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, ψ.

Step 1: Spot the isosceles triangle.

Segments BC and BD are both radii, so they have the same length. This means that CBD is isosceles, which also means that its base angles are congruent:
mC=mD=ψ

Step 2: Spot the straight angle.

Angle ABC is a straight angle, so
θ+mDBC=180mDBC=180θ

Step 3: Write an equation and solve for ψ.

The interior angles of CBD are ψ, ψ, and (180θ), and we know that the interior angles of any triangle sum to 180.
ψ+ψ+(180θ)=1802ψ+180θ=1802ψθ=02ψ=θ
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, ψ.

Step 1: Get clever and draw the diameter

Using the diameter, let's break ψ into ψ1 and ψ2 and θ into θ1 and θ2 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
(1)θ1=2ψ1
and
(2)θ2=2ψ2
because of what we learned in Case A.

Step 3: Add the equations.

θ1+θ2=2ψ1+2ψ2Add (1) and (2)(θ1+θ2)=2(ψ1+ψ2)Group variablesθ=2ψθ=θ1+θ2 and ψ=ψ1+ψ2
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: θ2 and ψ2 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:
(1)θ2=2ψ2
(2)(θ2+θ)=2(ψ2+ψ)

Step 3: Substitute and simplify.

(θ2+θ)=2(ψ2+ψ)(2)(2ψ2+θ)=2(ψ2+ψ)θ2=2ψ22ψ2+θ=2ψ2+2ψθ=2ψ
And we're done! We proved that θ=2ψ 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 ψ and θ. With a little algebra, we proved that θ=2ψ.
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 θ=2ψ.

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