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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.
A circle with three points on it. The second point is less than ninety degrees from the first point. The third point is more than one hundred eighty degrees from the second point. An arc made by the first and second point is labeled alpha. The angle made by the first point, the center, and the second point make an angle labeled theta. The angle made by the first point, the third point, and the second point is labeled psi.

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

What we're about to prove

A circle with three points on it. The second point is less than ninety degrees from the first point. The third point is more than one hundred eighty degrees from the second point. An arc made by the first and second point is labeled alpha. The angle made by the first point, the center, and the second point make an angle measuring fifty degrees. The angle made by the first point, the third point, and the second point is measured twenty-five degrees.
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
Case A has three points on the circle. The second point is less than ninety degrees clockwise from the first point. The third point is one hundred eighty degrees clockwise from the first point. The angle made by the first point, the center, and the second point is labeled theta. The angle made by the first point, the third point which passes through the center, and the second point is labeled psi.
In Case B, there are three points on the circle. The second point is more than ninety degrees clockwise from the first point. The third point is more than one hundred eighty degrees clockwise from the first point. The angle made by the first point, the center, and the second point is labeled theta. The angle made by the first point, the third point, and the second point is labeled psi.
In Case C there are three points on the circle. The second point is less than ninety degrees clockwise from the first point. The third point is less than one hundred eighty degrees clockwise from the first point. The angle made from the first point, the center, and the second point is labeled theta. The angle made from the first point, the third point, and the second point is labeled psi.
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, start color #11accd, \psi, end color #11accd.

Case A has three points on the circle. The second point is less than ninety degrees clockwise from the first point. The third point is one hundred eighty degrees clockwise from the first point. The angle made by the first point, the center, and the second point is labeled theta. The angle made by the first point, the third point which passes through the center, and the second point is labeled psi.

Step 1: Spot the isosceles triangle.

Three points A, C, and D are on the circle centered around point B. Point D is less than ninety degrees clockwise from the point A. The point C is one hundred eighty degrees clockwise from the point A. Line segment A C is a diameter. Line segment D C is a chord. Line segments B A, B C, and B D are radii that are a length of r units. The angle made by points A, B, and D are labeled theta. The angle made by points B C D is labeled psi. An angle made by points B D and C is labeled psi.
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.

Three points A, C, and D are on the circle centered around point B. Point D is less than ninety degrees clockwise from the point A. The point C is one hundred eighty degrees clockwise from the point A. Line segment A C is a diameter. Line segment D C is a chord. Line segments B A, B C, and B D are radii that are a length of r units. The angle made by points A, B, and D are labeled theta. The angle made by points B C D is labeled psi. An angle made by points B D and C is labeled psi. Angle C B D is labeled one hundred eighty degrees minus theta.
Angle angle, start color #e84d39, A, B, C, end color #e84d39 is a straight angle, so
θ+mDBC=180mDBC=180θ\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 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.
ψ+ψ+(180θ)=1802ψ+180θ=1802ψθ=02ψ=θ\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, start color #11accd, \psi, end color #11accd.

In Case B, there are three points on the circle. The second point is more than ninety degrees clockwise from the first point. The third point is more than one hundred eighty degrees clockwise from the first point. The angle made by the first point, the center, and the second point is labeled theta. The angle made by the first point, the third point, and the second point is labeled psi.

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:
In Case B, there are three points on the circle. The second point is more than ninety degrees clockwise from the first point. The third point is more than one hundred eighty degrees clockwise from the first point. The angle made by the first point, the center, and the second point is labeled theta. The angle made by the first point, the third point, and the second point is labeled psi. A point is also on the circle to make a line segment that passes through the center to the third point as a diameter. Angle theta one is on the left and theta two is on the right of the diameter where theta was located. Angle psi one is on the left and angle psi two is on the right of the diameter located where psi was.

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.

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

In Case C there are three points on the circle. The second point is less than ninety degrees clockwise from the first point. The third point is less than one hundred eighty degrees clockwise from the first point. The angle made from the first point, the center, and the second point is labeled theta. The angle made from the first point, the third point, and the second point is labeled psi.

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:
There are three points on the circle. The second point is less than ninety degrees clockwise from the first point. The third point is less than one hundred eighty degrees clockwise from the first point. The angle made from the first point, the center, and the second point is labeled theta. The angle made from the first point, the third point, and the second point is labeled psi. A point is on the circle with a line segment connecting it though the center to the third point making a diameter. The angle from the new point to the center to the first point is labeled theta two. The angle made by the center point, the third point, and the first point is labeled psi two.

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.

(θ2+θ)=2(ψ2+ψ)(2)(2ψ2+θ)=2(ψ2+ψ)θ2=2ψ22ψ2+θ=2ψ2+2ψθ=2ψ\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
Case A has three points on the circle. The second point is less than ninety degrees clockwise from the first point. The third point is one hundred eighty degrees clockwise from the first point. The angle made by the first point, the center, and the second point is labeled theta. The angle made by the first point, the third point which passes through the center, and the second point is labeled psi.
In Case B, there are three points on the circle. The second point is more than ninety degrees clockwise from the first point. The third point is more than one hundred eighty degrees clockwise from the first point. The angle made by the first point, the center, and the second point is labeled theta. The angle made by the first point, the third point, and the second point is labeled psi.
In Case C there are three points on the circle. The second point is less than ninety degrees clockwise from the first point. The third point is less than one hundred eighty degrees clockwise from the first point. The angle made from the first point, the center, and the second point is labeled theta. The angle made from the first point, the third point, and the second point is labeled psi.
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
In Case B, there are three points on the circle. The second point is more than ninety degrees clockwise from the first point. The third point is more than one hundred eighty degrees clockwise from the first point. The angle made by the first point, the center, and the second point is labeled theta. The angle made by the first point, the third point, and the second point is labeled psi. A point is also on the circle to make a line segment that passes through the center to the third point as a diameter. Angle theta one is on the left and theta two is on the right of the diameter where theta was located. Angle psi one is on the left and angle psi two is on the right of the diameter located where psi was.
There are three points on the circle. The second point is less than ninety degrees clockwise from the first point. The third point is less than one hundred eighty degrees clockwise from the first point. The angle made from the first point, the center, and the second point is labeled theta. The angle made from the first point, the third point, and the second point is labeled psi. A point is on the circle with a line segment connecting it though the center to the third point making a diameter. The angle from the new point to the center to the first point is labeled theta two. The angle made by the center point, the third point, and the first point is labeled psi two.
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.

Want to join the conversation?

  • leaf green style avatar for user Pranav
    I need help in the proofs for Case 3 in inscribed angles
    (9 votes)
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  • male robot johnny style avatar for user toma.gevorkyan8
    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 votes)
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  • blobby green style avatar for user Akira
    What happens to the measure of the inscribed angle when its vertex is on the arc? Will it be covered in the future lecture?
    (4 votes)
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    • blobby green style avatar for user Reynard Seow
      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).
      (0 votes)
  • blobby green style avatar for user Konstantin Zaytsev
    Why do you write m in front of the angle sign?
    (1 vote)
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  • blobby green style avatar for user Jason Showalter
    What is the greatest measure possible of an inscribed angle of a circle?
    (2 votes)
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    • piceratops ultimate style avatar for user Pat Florence
      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
      (2 votes)
  • blobby green style avatar for user avery.jackson
    I don't understand was a radian angle is and how to get the circumference from it. I also mess up when fractions and the pie symbol are used.
    (2 votes)
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    • duskpin ultimate style avatar for user victoriamathew12345
      The radians for an angle are based on how many radii equal the length of the same arc subtended by that angle. In relation to the circumference, the circumference is equal to 2(pi)(r) r meaning radius, not radians (there is a difference). The circumference can also be seen as the arc for the whole circle and in an arc there are 2 pi radii, so there are 2 pi radians in a whole entire circle.
      (2 votes)
  • duskpin seedling style avatar for user Trinity Kelly
    Ok so I have a small question, I'm doing something called VLA and they gave me two different equations one to find the radius using the circumference, and the other to find the diameter also using the circumference, the equations were. Circumference/p = diameter, and the other was circumference/2p = radius, but i'm confused cause when I used the second one, it would give me a really big number while the first equation gave me a smaller number. Also sorry if this has nothing to do with what you were talking about Sal, I was waiting until I had enough energy to be able to ask my question.
    (1 vote)
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  • winston default style avatar for user fluffyplatypus
    So for the central angle to be double of the inscribed angle, the rays of the inscribed angle should originate from the point of intersection of the points (on the circumference of the circle) of the central angle?
    (2 votes)
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  • duskpin ultimate style avatar for user taylor k.
    Do all questions have the lines colored? If not, how would you distinguish between the two?
    (1 vote)
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  • aqualine ultimate style avatar for user Simum
    Why is (θ2+θ)=2(ψ2+ψ)?
    (1 vote)
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