In Figure 1, circle *O* has radii OA, OB, OC and OD If chords AB and CD are of equal length, it can be shown that Δ AOB ≅ Δ DOC. This would make *m* ∠1 = *m* ∠2, which in turn would make *m* *m*

**Figure 1 **A circle with four radii and two chords drawn.

*Theorem 78:* In a circle, if two chords are equal in measure, then their corresponding minor arcs are equal in measure.

The converse of this theorem is also true.

*Theorem 79:* In a circle, if two minor arcs are equal in measure, then their corresponding chords are equal in measure.

**Example 1:** Use Figure 2 to determine the following. (a) If *AB = CD*, and *m* CD. (b) If *m*

**Figure 2 **The relationship between equality of the measures of (nondiameter) chords and equality of the measures of their corresponding minor arcs.

*m* *(Theorem 78)*

*GH* = 8 *(Theorem 79)*

Some additional theorems about chords in a circle are presented below without explanation. These theorems can be used to solve many types of problems.

*Theorem 80:* If a diameter is perpendicular to a chord, then it bisects the chord and its arcs.

In Figure 3, UT, diameter QS is perpendicular to chord QS By *Theorem 80, QR = RS, m* *m* *m* *m*

**Figure 3** A diameter that is perpendicular to a chord.

*Theorem 81:* In a circle, if two chords are equal in measure, then they are equidistant from the center.

In Figure 4, if *AB = CD*, then by *Theorem 81, OX = OY*.

**Figure 4 **In a circle, the relationship between two chords being equal in measure and being equidistant from the center.

*Theorem 82:* In a circle, if two chords are equidistant from the center of a circle, then the two chords are equal in measure.

In Figure 5, if *OX* = *OY*, then by *Theorem 82, AB = CD*.

**Example 2:** Use Figure to find *x*.

**Figure 5 **A circle with two minor arcs equal in measure.

**Example 3:** Use Figure 6, in which *m* *m* *AC*.

**Figure 6 **A circle with two minor arcs equal in measure.

**Example 4:** Use Figure 7, in which *AB* = 10, *OA* = 13, and *m* ∠ *AOB* = 55°, to find *OM*, *m* *m*

**Figure 7** A circle with a diameter perpendicular to a chord.

So, ST ⊥ AB, and ST is a diameter. *Theorem 80* says that *AM = BM*. Since *AB* = 10, then *AM* = 5. Now consider right triangle *AMO*. Since *OA* = 13 and *AM* = 5, *OM* can be found by using the Pythagorean Theorem.

Also, *Theorem 80* says that *m* *m* *m* *m* *m* ∠ *AOB* = 55°, that would make *m* *m* *m* *m*

**Example 5:** Use Figure 8, in which *AB* = 8, *CD* = 8, and *OA* = 5, to find *ON*.

**Figure 8 **A circle with two chords equal in measure.

By *Theorem 81*, *ON* = *OM*. By *Theorem 80*, *AM = MB*, so AM = 4. *OM* can now be found by the use of the *Pythagorean Theorem* or by recognizing a Pythagorean triple. In either case, *OM* = 3. Therefore, *ON* = 3.