# Circles

### Topics

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 ⊙*O'* and *AB* is congruent to *CD*, we need to prove that ∠*AOB* is congruent to ∠*CO'D*. Because *OA*, *OB*, *O'C*, and *O'D* 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 Δ*AOB* is congruent to Δ*CO'D* by SSS. Therefore we also know that ∠*AOB* is congruent to ∠*CO'D*.

Congruent triangles can help us out in proving the second part, too.

For the two congruent circles ⊙*O* and ⊙*O'*, we already know ∠*AOB* is congruent to ∠*CO'D*. Just the same, because *OA*, *OB*, *O'C*, and *O'D* 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 ∠*CO'D*, we know that Δ*AOB* is congruent to Δ*CO'D* by SAS. We know that *AB* is congruent to *CD*, because corresponding pieces of congruent triangles are congruent. That wasn't too bad, was it?