Look at the circle below. I randomly placed points A, B, and C, and the center is marked O. I also marked the angle ACB.

First of all, if you move the point C to wherever you want on the circle, the angle ACB will still have the same measure. The reason of this will become clear after I prove the central angle theorem.

The central angle theorem states that angle AOB is always double of ACB, regardless of the placement of the points A, B, and C.

The proof of this will take three steps. For the first step, I have moved the point C so that the line AC goes through O. Basically, AC is the diameter of the circle.

I have marked the angle ACB with a thick line and angle AOB with a thin line. There is also a dotted line to further identify angles.

First, we must take note of the fact that since O is the center of the circle, any line containing O in it is a radius of the circle. So, the line AO, BO, and CO are all radii of the circle. Since the radius is always the same measurement, AO, BO, and CO are congruent. The tick marks on these three lines are a way of saying that all lines with that mark are congruent.

Now, we are going to use some triangle theorems. I don't have time to prove them both now, but I'm sure you could prove them pretty easily.

Isosceles Triangle Theorem: If two sides of a triangle are congruent, their opposite angles are congruent. In this example, since BO and OC are congruent, the angles BCO and OBC are congruent.

Exterior Angle Theorem: If you have a triangle where an exterior angle is shown, its measurement is the sum of the two remote interior angles. For instance, in the triangle below, angle d is the sum of angle a and angle c. In the example above, angle AOB is the sum of angles BCO and OBC.

Because of the isosceles angle theorem, we already determined that angle BCO is congruent to angle OBC. Rather than just marking them congruent, let's let x be their measurement.

Now, we will use the exterior angle theorem. We already determined that angle AOB is the sum of angles BCO and OBC. And, we already determined that these two angles have a measure of x. So, we can determine the measure of angle AOB.

m<AOB = m<BCO + m<OBC

m<AOB = x + x

m<AOB = 2x

So, angle AOB has a measure of 2x.

The central angle theorem says that the measure of angle AOB is double of the measure of angle ACB. As you can see, m<AOB = 2x and m<ACB = x, so we have proved the first part.

Now, imagine that C was not on the diameter. Say you had an image like the following:

In this example, I have kept the information that we already found. The angles that are thick lines are the ones we are testing. Before we continue, we must first identify a few angle measurements. All of them will be found using the same theorems as before. I have made the angle measures blue so it is easier to keep track.

Also, note the fact that a triangle's angles always add up to 180°.

If you look at triangle AOB, you will see that all three angles have a measurement. We know that if we add these up, we will have to get 180.

2x + y + y = 180

2x + 2y = 180

This fact will definitely help in the future. Also, we should look at the fact that BO and OC both have a tick mark, meaning that they are congruent. Because of the isosceles triangle theorem, we know that <BCO and <OBC are congruent.

In triangle BOC, we have a few of the angle measurements. One of them is 2y, which is already a part of the 180° in the triangle. We also have a 2z, which is not already a part of it. So, we will must add 2z to the equation. But, when you add 2z, you must also subtract back out a 2z. So, it will look like this:

2x + 2y = 180

2x + 2y + 2z - 2z = 180

2y + 2z + 2x - 2z = 180

(2y + 2z) + (2x - 2z) = 180

We have now separated what we know already from the new information, which is that the rest of the angles add up to 2x - 2z. But, we also determined that they are congruent. So, we will just divide 2x - 2z by two to find out these angle measurements.

(2x - 2z) = 2(x - z)

So, this measurement is x - z.

But, this proof requires the measure of angle ACB. We almost have it. We found that it is made up of z and x-z.

z + x - z = x

So, m<ACB = x. Again, this is half of 2x, so we have proven this type of example.

We have proven it if C is on the diameter or past the diameter. But, what if it is before the diameter?

I have again labeled some angle measures using the isosceles triangle theorem and exterior angle theorem. This time, the new variable is n. Note that as before, 2x + 2y = 180.

Angle BAO has a measure of y and angle CAO has a measure of n. So, angle BAC should therefore have a measure of y-n.

So, let's see what measurements we have for triangle ABC. We have the y-n, the y, and the x, as well as two unidentified angles that must sum to 2x + 2y.

y - n + y + x + ___ = 2x + 2y

2y + x - n + ___ = 2x + 2y

___ = x + n

So, we must figure out how to split up the x + n between angles CB• and ACB.

If you look at triangle BOC, you will notice that there are two congruent sides! So, we can use the isosceles triangle theorem to say that <OBC is congruent to <OCB. <OBC has a part with x and <OCB has a part with n.

In order to make them equal, we must give <OBC the n and <OCB the x. This will make both measures x+n, which will allow them to be congruent.

And where does this x go? x is put as the measure of <ACB. Since <AOB is double of this, we have proved the last possibility.

These theorems and variables are pretty complicated. However, like I said in the beginning, it will make sense if you actually try to prove it yourself. Draw a circle and start trying to prove the different possibilities. This is how I was able to understand it. I hope this is a guide to try to prove it, but it may be difficult to understand without actually following along.

Thanks for this post – despite the convoluted diagrams, it was quite helpful!

ReplyDelete