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Inscribed Angle Theorem

Here's a triangle inscribed in a circle.

What makes it an "inscribed triangle"?
  1. It's a triangle.
  2. Each vertex is on the circle.
Use the Angle tool to measure and :
  1. Click the Angle tool, second from the right, to activate it. Its frame will change color to show that it's active.
  2. Click A, then B, then C. Each point will grow a "halo" to confirm your click. If you click but don't see the halo, try again.
  3. Click A, then D, then C. (Notice that the Angle tool stays active - so when you want to do something else, pick another tool.)
Use the Move tool to manipulate the diagram. As you do, compare the measures of and .
  1. Click the Move tool, leftmost on the toolbar, to activate it. Its frame will change color to show that it's active.
  2. Click and drag A, B, and/or C. Look for relationships between the angle measures that you see.

What do you notice about the measures of and ?

The Inscribed Angle Theorem is observation #3 in my answer to the question above. Let's prove that it's always true. We'll look at three cases: Case 1: the center of the circle is on one of the rays (sides) of the inscribed angle. Case 2: the center of the circle is in the interior of the inscribed angle. Case 3: the center of the circle is in the exterior of the inscribed angle.

Case 1: D is on a side of the angle.

In the figure below, I've shown the situation where D is on ray BA. The thinking goes the same way if D is on ray BC instead.

Do you see any isosceles triangles? How do you know they are isosceles?

Where do isosceles triangles give us congruent angles?

Same info another way: matching angles are congruent in the figure below.

In the figure above, there are (at least) two ways to add onto the teal angle () to make . Which do you see?

Since , we conclude that . (This is one of the Exterior Angle theorems: for any triangle, one exterior angle equals the sum of the "other two" interior angles.) Since , we conclude that . But , so . This proves Case 1.

Case 2: D is between ray BA and ray BC.

Line BE cuts the circle into two halves: one with A; the other with C. Move A and C freely within their own halves. You can move B if you like, but make sure that line BE never passes through A or C.

We can apply Case 1 to the tangerine angles. It tells us that . What does Case 1 tell us about the green angles?

How are the purple angles related?

Case 3: D is neither *on* nor *between* rays BA and BC.

Let's reuse the Case 2 figure. Move A, B, and C so that they are still in clockwise order but A and C are on the same side of line BE; that is, line BE does not meet segment .

How are the purple angles related?

I'm putting a link here to a very surprising 5-minute 3B1B video: https://www.youtube.com/watch?v=HEfHFsfGXjs. The sequel to that video uses the Inscribed Angle Theorem to explain what happens in the first video.