# Circles

### Topics

## Introduction to :

There seems to be an awful lot of possibilities for inscribed angles. Our old buddies, central angles, always had their vertices at the center. But an inscribed angle's vertex can be anywhere on the circle. Is the center of the circle even relevant anymore?

It turns out the center of the circle is very relevant to inscribed angles. The center can help us get organized here. Let's classify all possible central angles ∠*ABC* based on whether the center *O* of the circle is "inside," "outside," or "on" the angle, and examine these brief cases.

### Case 1

*O* is on one of the chords which make up ∠*ABC*. In other words, one of the sides of ∠*ABC* is a diameter of ⊙*O*.

Let *x* stand for the measure of the inscribed angle ∠*ABC*, and we'll see if we can find m*AC*. We'll draw in line segment *OA*, because if we can find the measure of the central angle ∠*AOC*, we'll know m*AC*.

By drawing in *OA*, we've made Δ*AOB*, which must be an isosceles triangle since *OB* and *OA* are radii of the same circle and are therefore congruent. That means m∠*BAO* = m∠*ABC* = *x*. We've also conveniently made ∠*AOC* an exterior angle, which means that m∠*AOC* = m∠*BAO* + m∠*ABC* = *x* + *x* = 2*x*. So m*AC* = 2*x*.

Now we know that a Case 1 angle intercepts an arc with twice the measure of the angle. We can use that knowledge to examine Case 2.

### Case 2

*O* is in the interior of ∠*ABC*.

Let's add a diameter in there.

Now we have two Case 1 angles ∠ABD and ∠DBC. From our work on Case 1, we can say that m∠*ABD* = ½ × m*AD* and m∠*DBC* = ½ × m*DC*

Since ∠*ABD* and ∠*DBC* are adjacent angles, m∠*ABC* = m∠*ABD* + m∠*DBC*. Substituting, we have m∠*ABC* = ½ × m*AD* + ½ × m*DC*. Factoring out the common ½, we have m∠*ABC* = ½ × (m*AD* + m*DC*).

Now we bring in our Arc Addition Postulate. Seemed silly at the time, didn't it? Well, it's here to save us now.

m*AD* + m*DC* = m*AC*

Now we substitute *that* in, and we have m∠*ABC* = ½ × m*AC*.

In other words, a Case 2 inscribed angle intercepts an arc with twice the measure of the angle, just as a Case 1 angle does. Curious. Now let's look at Case 3.

### Case 3

*O* is in the exterior of ∠*ABC*.

Diameters have been good to us in the past, so how about we throw one in there, just to see what happens? After all, what's the worst that could happen? It's just math.

We can actually use the same strategy we used in Case 2 with a little twist. Here, we have two Case 1 angles, ∠*CBD*, which intercepts arc *CD*, and ∠*ABD*, which intercepts arc *ACD*. From our work in Case 1, we know that m∠*ABD* = ½ × m*ACD* and m∠*CBD* = ½ × m*CD*.

We'll find our target angle by subtraction this time, not by addition (although you can't have subtraction without addition, so in a way, we're still using addition).

If we subtract, we get m∠*ABC* = m∠*ABD* – m∠*CBD*. Substituting, we have m∠*ABC* = ½ × m*ACD* – ½ × m*CD*. Factoring, we have m∠*ABC* = ½ × (m*ACD* – m*CD*).

Now we bring in our old friend, the Arc Addition Postulate, and we get m∠*ABC* = ½ × m*AC*.

We've shown that a Case 3 inscribed angle intercepts an arc with twice the measure of the angle—same as a Case 1 angle or a Case 2 angle. Since any inscribed angle falls into one of the three cases, we have proved the **Inscribed Angle Theorem**: for ∠*ABC* inscribed in a circle containing points *A* and *C*: m*AC* = 2 × m∠*ABC*.

### Sample Problem

Given that ∠*ACB* is inscribed in a circle containing points *A* and *B*, and that m∠*ACB* = 42°, what is the measure of arc *AB*?

Here we have an inscribed angle intercepting an arc, so we can use the Inscribed Angle Theorem. We can start out with m*AB* = 2 × m∠*ACB*, and then substitute 42° for m∠*ACB*. That will give us m*AB* = 2 × 42°, or m*AB* = 84°.

### Sample Problem

Given that ∠*ACB* is inscribed in circle ⊙*O* with radius 5 cm containing points *A* and *B*, and that m∠*ACB* = 78°, what is the length of arc *AB*?

Before we calculate arc length, we can calculate the measure of the arc using the Inscribed Angle Theorem. We can calculate m*AB* to be 2 × m∠*ACB* = 2 × 78° = 156°. We can use the arc length formula , with *θ* = 156° and *r* = 5 cm. This should give us an arc length of about 13.6 cm.