Tired of ads?
Join today and never see them again.
Advertisement - Guide continues below
Fun activity time! You'll need a hula-hoop and several huge rubber bands big enough to stretch all the way across the hula-hoop. If you don't have those things, that's fine too. You can just imagine them. Usually you have to buy those industrial-sized rubber bands by the pallet.
Given those materials, how would you go about building a musical instrument? Banging the hula-hoop on the driveway and calling it a "percussion instrument" doesn't count.
You could build a very simple harp or lyre-like instrument by stringing a rubber band across the center of the circle, like this:
Pluck that rubber band, and you've got a note. Congratulations! You can now play any piece of music ever written…as long as it's Mozart's Monotone Concerto…which doesn't actually exist.
If you want to play more than one pitch, you'll need more rubber bands. You'll have to stretch them across the circle at different places so that the rubber bands end up being stretched to different lengths. Like this.
In geometry, we call objects like these rubber bands chords. No, we're not lyre-ing.
A line segment is a chord of ⊙O if both the segment's endpoints are on ⊙O. A chord that goes through the center of the circle is special, so we give it a special name: the diameter. A chord of ⊙O is a diameter of ⊙O if it contains O.
The length of a diameter of a circle is called "the diameter of the circle," just as the length of a radius of a circle is called "the radius of the circle." Hopefully that doesn't annoy you. If it does, blame the ancient Greeks. Yes, we know that's a bit of a cop-out. Blame the ancient Greeks for that too. And of course, we already know from basic geometry that a circle's diameter is exactly twice its radius. Small world, right?
How many chords are in the following figure? How many diameters?
First, let's find all the chords. Any segment whose endpoints are on ⊙O are chords. Just looking at the figure, we can see that we have 5 chords: AG, AE, BF, CG, and DF. Diameters are chords that go through the center, which means we can have a maximum of 5 diameters. In this case, we only have 2: AE and BF. While OC goes through point O, it is not a diameter because only one of its endpoints is on ⊙O (the other is O, the center, which makes it a radius).
If you think about it, circles are all about friendship. We saw before that every central angle has a buddy, and every arc has a buddy. Plus, where do you think the phrase "inner circle" came from?
Unfortunately, chords don't have built-in buddies of their own kind, but they are on good terms with central angles. So even if they aren't that close, they're still chord-ial.
A chord is defined by a circle and two points on that circle, remember? Here, we have a chord defined by ⊙O and points A and B on ⊙O.
We can use the same elements that define a chord to define a central angle. See? Just draw a couple of line segments from A to O and B to O and you've got a nice, shiny central angle: ∠AOB.
A chord and its central angle aren't just acquaintances. In fact, they have a very close relationship. In congruent circles, two chords are congruent if and only if their associated central angles are congruent. We'll call this the Angle-Chord Theorem.
Another biconditional. Like all biconditionals, we can break this one down into its two parts. First, if two chords are congruent, then their associated central angles are congruent. Second, if two central angles are congruent, then their associated chords are congruent.
To prove this, we're going to take advantage of the fact that a chord and its associated central angle form a triangle. Remember triangles, our old pals? Well, they're here to help us now.
Given that ⊙O is congruent to itself and AB is congruent to CD, we need to prove that ∠AOB is congruent to ∠COD. Because OA, OB, OC, and OD are all radii of congruent circles, they must all be congruent by definition of congruent circles. Since we are also given that AB is congruent to CD, we know that ΔAOB is congruent to ΔCOD by SSS. Therefore we also know that ∠AOB is congruent to ∠COD.
Congruent triangles can help us out in proving the second part, too.
We already know ∠AOB is congruent to ∠COD. Just the same, because OA, OB, OC, and OD are all radii of congruent circles, they must all be congruent by definition of congruent circles. Since we are also given that ∠AOB is congruent to ∠COD, we know that ΔAOB is congruent to ΔCOD by SAS. That means AB is congruent to CD, because corresponding pieces of congruent triangles are congruent. That wasn't too bad, was it?
Speaking of friends, let's return (notice we refrained from saying "circle back") to our discussion of friendships. We know that arcs are great friends with central angles. We just learned that chords are also great friends with central angles. It's not too far-fetched to imagine a central angle throwing a party and inviting both its associated chord and its associated arc. Who's bringing the hummus?
Indeed, since each central angle "knows" an arc and a chord, it can form the bridge of friendship between those two—introducing them to each other, making sure they have plenty to talk about, overall making sure they get to know each other and are having a good time at the party.
An arc and a chord that share a central angle ought to get along just fine. After all, they have two points in common.
In fancy talk, two chords are congruent if and only if their associated arcs are congruent. We've got another biconditional here, and you know what that means: we have to prove both directions of the statement.
In one way, if two chords are congruent, then their associated arcs are congruent. In the other, if two arcs are congruent, then their associated chords are congruent.
Given that ⊙O is congruent to ⊙O' with chords AB and CD, we can start by drawing in some extra line segments: OA, OB, O'C, and O'D.
These segments form the central angles of chords AB and CD: ∠AOB and ∠CO'D, respectively. We are given that AB is congruent to CD. Since the chords are congruent, we know that ∠AOB and ∠CO'D are congruent as well. Since arcs AB and CD have the same measure and are segments of congruent circles, we can say that arcs AB and CD are congruent by definition of congruent arcs.
To prove that AB is congruent to CD given that arc AB is congruent to arc CD, we can use pretty much the same logic we used, but in reverse. Here, we're given that arcs AB and CD are congruent. Then by definition of congruent arcs, we know that their associated central angles, ∠AOB and ∠CO'D, are also congruent. That means chords AB and CD are congruent.
We'll call this relationship the Arc-Chord Theorem. And so begins a glorious friendship between the arc and the chord.