Inscribed angle theorem proof (article)

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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.

An angle is a $central angle$ ‍ when its vertex is the center of the circle, such as $θ$ ‍ shown in the image.

An angle is an $inscribed angle$ ‍ when its vertex is on the circle, such as $ψ$ ‍ shown in the image.

The rays of inscribed and central angles intercept part of the circumference of the circle. We call this part of the circumference $the intercepted arc$ ‍.

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

$θ = 2 ψ$ ‍

Proof overview

To prove $θ = 2 ψ$ ‍ for all $θ$ ‍ and $ψ$ ‍ (as we defined them above), we must consider three separate cases:

Case A	Case B	Case 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, $ψ$ ‍.

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 $\overset{―}{B C}$ ‍ and $\overset{―}{B D}$ ‍ are both radii, so they have the same length. This means that $△ C B D$ ‍ is isosceles, which also means that its base angles are congruent:

$m ∠ C = m ∠ D = ψ$ ‍

Step 2: Spot the straight angle.

Angle $∠ A B C$ ‍ is a straight angle, so

$\begin{aligned} θ + m ∠ D B C & = 180^{\circ} \\ m ∠ D B C & = 180^{\circ} - θ \end{aligned}$ ‍

Step 3: Write an equation and solve for $ψ$ ‍.

The interior angles of $△ C B D$ ‍ are $ψ$ ‍, $ψ$ ‍, and $(180^{\circ} - θ)$ ‍, and we know that the interior angles of any triangle sum to $180^{\circ}$ ‍.

$\begin{aligned} ψ + ψ + (180^{\circ} - θ) & = 180^{\circ} \\ 2 ψ + 180^{\circ} - θ & = 180^{\circ} \\ 2 ψ - θ & = 0 \\ 2 ψ & = θ \end{aligned}$ ‍

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, $ψ$ ‍.

Step 1: Get clever and draw the diameter

Using the diameter, let's break $ψ$ ‍ into $ψ_{1}$ ‍ and $ψ_{2}$ ‍ and $θ$ ‍ into $θ_{1}$ ‍ and $θ_{2}$ ‍ 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

Step 3: Add the equations.

$\begin{aligned} θ_{1} + θ_{2} & = 2 ψ_{1} + 2 ψ_{2} & Add (1) and (2) \\ (θ_{1} + θ_{2}) & = 2 (ψ_{1} + ψ_{2}) & Group variables \\ θ & = 2 ψ & θ = θ_{1} + θ_{2} and ψ = ψ_{1} + ψ_{2} \end{aligned}$ ‍

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: $θ_{2}$ ‍ and $ψ_{2}$ ‍ 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:

$(1) θ_{2} = 2 ψ_{2}$ ‍

$(2) (θ_{2} + θ) = 2 (ψ_{2} + ψ)$ ‍

Step 3: Substitute and simplify.

$\begin{aligned} (θ_{2} + θ) & = 2 (ψ_{2} + ψ) & (2) \\ (2 ψ_{2} + θ) & = 2 (ψ_{2} + ψ) & θ_{2} = 2 ψ_{2} \\ 2 ψ_{2} + θ & = 2 ψ_{2} + 2 ψ \\ θ & = 2 ψ \end{aligned}$ ‍

And we're done! We proved that $θ = 2 ψ$ ‍ 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 A	Case B	Case 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 $ψ$ ‍ and $θ$ ‍. With a little algebra, we proved that $θ = 2 ψ$ ‍.

In cases B and C, we cleverly introduced the diameter:

Case B	Case 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.

Case B

Case 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 $θ = 2 ψ$ ‍.

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Pranav
5 years agoPosted 5 years ago. Direct link to Pranav's post “I need help in the proofs...”
I need help in the proofs for Case 3 in inscribed angles
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(15 votes)
toma.gevorkyan8
7 years agoPosted 7 years ago. Direct link to toma.gevorkyan8's post “Hi Sal, I have a question...”
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....
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(8 votes)
- gavinjanz24
  2 years agoPosted 2 years ago. Direct link to gavinjanz24's post “5 years later... I wonder...”
  See Also
  Inscribed angles (video) | Circles | Khan Academy
  5 years later... I wonder if Sal is still working on it.
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  (10 votes)
kjohnson8937
a year agoPosted a year ago. Direct link to kjohnson8937's post “can I use ψ as a variable...”
can I use ψ as a variable to measure any angle I want to?
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(4 votes)
- kubleeka
  a year agoPosted a year ago. Direct link to kubleeka's post “Yes, and it doesn't have ...”
  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.
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pandabuff2016
10 months agoPosted 10 months ago. Direct link to pandabuff2016's post “is it possible to prove c...”
is it possible to prove case c without proving a & b first?
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(5 votes)
- jonhlhn.surf
  9 months agoPosted 9 months ago. Direct link to jonhlhn.surf's post “You do not need to prove ...”
  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.
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Jason Showalter
4 years agoPosted 4 years ago. Direct link to Jason Showalter's post “What is the greatest meas...”
What is the greatest measure possible of an inscribed angle of a circle?
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(4 votes)
- Pat Florence
  4 years agoPosted 4 years ago. Direct link to Pat Florence's post “If the angle were 180, th...”
  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
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  (6 votes)
Akira
3 years agoPosted 3 years ago. Direct link to Akira's post “What happens to the measu...”
What happens to the measure of the inscribed angle when its vertex is on the arc? Will it be covered in the future lecture?
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- Reynard Seow
  2 years agoPosted 2 years ago. Direct link to Reynard Seow's post “If the vertex of the insc...”
  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).
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taylor k.
4 years agoPosted 4 years ago. Direct link to taylor k.'s post “Do all questions have the...”
Do all questions have the lines colored? If not, how would you distinguish between the two?
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- victoriamathew12345
  3 years agoPosted 3 years ago. Direct link to victoriamathew12345's post “Normally, to distinguish ...”
  Normally, to distinguish between two lines, you would have letters instead.
  E.g: f(x) vs g(x)
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Konstantin Zaytsev
4 years agoPosted 4 years ago. Direct link to Konstantin Zaytsev's post “Why do you write m in fro...”
Why do you write m in front of the angle sign?
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- KC
  4 years agoPosted 4 years ago. Direct link to KC's post “m=measure so it would jus...”
  m=measure so it would just be the measure of the angle
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eperez3463
a year agoPosted a year ago. Direct link to eperez3463's post “how can i solve this”
how can i solve this
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Trinity Kelly
5 years agoPosted 5 years ago. Direct link to Trinity Kelly's post “Ok so I have a small ques...”
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.
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- kubleeka
  5 years agoPosted 5 years ago. Direct link to kubleeka's post “When you compute C/2π, be...”
  When you compute C/2π, be sure that you're dividing by π by putting the denominator in parentheses. If you just enter C/2*π, the calculator will follow order of operations, computing C/2, then multiplying the result by π.
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Inscribed angle theorem proof (article) | Khan Academy (2024)

FAQs

How do you prove the inscribed angle theorem? ›

Inscribed angle is between a chord and the diameter of a circle. In the above image, let us consider that ∆OBD is an isosceles triangle where OD = OB = radius of the circle. Therefore, ∠ODB = ∠DBO = inscribed angle = θ. The diameter AD is a straight line hence ∠BOD = 180 - ∠AOB(call it x).

Read The Full Story ›

How to solve inscribed angle in geometry? ›

Step 1: Identify the intercepted arc of the central angle and the intercepted arc of the inscribed or circ*mscribed angle. Ensure they are the same. Step 2: For an inscribed angle, the measure of the angle is one-half of the measure of the central angle.

Discover More ›

What is the inscribed angles of a circle theorem? ›

The inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle. Therefore, the angle does not change as its vertex is moved to different positions on the circle.

What is the measure of the inscribed angle to the arc length? ›

The Inscribed Angle Theorem states that the measure of an inscribed angle is half the measure of its intercepted arc. Inscribed angles that intercept the same arc are congruent. This is called the Congruent Inscribed Angles Theorem and is shown below.

Learn More Now ›

What are the 4 theorems on inscribed angles? ›

Inscribed Angles Intercepting Arcs Theorem

Inscribed angles that intercept the same arc are congruent. Angles Inscribed in a Semicircle Theorem Angles inscribed in a semicircle are right angles. Cyclic Quadrilateral Theorem The opposite angles of a cyclic quadrilateral are supplementary.

Keep Reading ›

What is the proof of the angle angle theorem? ›

Angle Angle Side Congruence Theorem

If both the triangles are superimposed on each other, we see that ∠B =∠E and ∠C =∠F. And the non-included sides AB and DE are equal in length. Therefore, we can say that ∆ABC ≅ ∆DEF.

Read On ›

What is the formula for inscribed angle? ›

Inscribed Angle Theorem:

The measure of an inscribed angle is half the measure of the intercepted arc. That is, m ∠ A B C = 1 2 m ∠ A O C . This leads to the corollary that in a circle any two inscribed angles with the same intercepted arcs are congruent.

Discover More Details ›

Why are inscribed angles half the arc? ›

Because a semicircle (half a circle) creates an intercepted arc that measures 180°, therefore, any corresponding inscribed angle would measure half of it, as Varsity Tutors nicely states.

Explore More ›

How will you know that an inscribed angle is a right angle? ›

Corollary (Inscribed Angles Conjecture III ): Any angle inscribed in a semi-circle is a right angle. Proof: The intercepted arc for an angle inscribed in a semi-circle is 180 degrees. Therefore the measure of the angle must be half of 180, or 90 degrees. In other words, the angle is a right angle.

Keep Reading ›

How does the measure of an inscribed angle compare to the measure of the arc it intercepts? ›

An inscribed angle is formed when two lines pass through the circle's circumference and meet at a vertex on another part of the circle's circumference. The intercepted arc that is formed is equal to the inscribed angle, multiplied by two (intercepted arc measure = inscribed angle * 2).

How to find arc length of a circle without central angle? ›

The length of the arc without using the central angle can be determined by the given method. Step 1: Multiply the sector area of the given circle by 2. Step 2: Divide the number by the square of the radius.

Get More Info Here ›

How do you prove the inscribed quadrilateral theorem? ›

Proof: In the quadrilateral ABCD can be inscribed in a circle, then we have seen above using the inscribed angle theorem that the sum of either pair of opposite angles = (1/2(a1 + a2 + a3 + a4) = (1/2)360 = 180. Conversely, if the quadrilateral cannot be inscribed, this means that D is not on the circumcircle of ABC.

How do you prove the angle angle side theorem? ›

In order to use AAS, all that is necessary is identifying two equal angles in a triangle, then finding a third side adjacent to only one of the angles in each of the triangles such that the two sides are equal. This is enough to prove the two triangles are congruent.

How do you prove the corresponding angle theorem? ›

How do you prove the Corresponding Angles Theorem? Let there be two parallel lines crossed by a transversal forming an angle, a, and an adjacent angle, b, that is below it. These two angles must be supplementary since they form a straight angle.

Explore More ›

How do you prove the angle sum theorem? ›

We can draw a line parallel to the base of any triangle through its third vertex. Then we use transversals, vertical angles, and corresponding angles to rearrange those angle measures into a straight line, proving that they must add up to 180°.

Get More Info ›

Inscribed angle theorem proof (article) | Khan Academy (2024)

Getting started

What we're about to prove

Proof overview

Case A: The diameter lies along one ray of the inscribed angle, ψ‍.

Step 1: Spot the isosceles triangle.

Step 2: Spot the straight angle.

Step 3: Write an equation and solve for ψ‍.

Case B: The diameter is between the rays of the inscribed angle, ψ‍.

Step 1: Get clever and draw the diameter

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

Step 3: Add the equations.

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

Step 1: Get clever and draw the diameter

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

Step 3: Substitute and simplify.

A summary of what we did

Want to join the conversation?

FAQs

How do you prove the inscribed angle theorem? ›

How to find arc length of a circle without central angle? ›

Case A: The diameter lies along one ray of the inscribed angle, $ψ$ ‍.

Step 3: Write an equation and solve for $ψ$ ‍.

Case B: The diameter is between the rays of the inscribed angle, $ψ$ ‍.