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 
. This is stated as a theorem.
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| Figure 1 | A circle with four radii and two chords drawn. | |
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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 = 60°, find m CD. (b) If m 
= 
and EF = 8, find GH.
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| Figure 2 | The relationship between equality of the measures of (nondiameter) chords and equality of the measures of their corresponding minor arcs. | |
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m = 60° (Theorem 78)
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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 
, and m 
= m 
.
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| Figure 3 | A diameter that is perpendicular to a chord. | |
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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.
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| Figure 4 | In a circle, the relationship between two chords being equal in measure and being equidistant from the center. | |
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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 4 , if OX = OY, then by Theorem 82, AB = CD.
Example 2: Use Figure 5 to find x.
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| Figure 5 | A circle with two minor arcs equal in measure. | |
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Example 3: Use Figure 6 , in which m 
= 115°, m 
= 115°, and BD = 10, to find AC.
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| Figure 6 | A circle with two minor arcs equal in measure. | |
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Example 4: Use Figure 7 , in which AB = 10, OA = 13, and m ∠ AOB = 55°, to find OM, m 
and m 
.
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| Figure 7 | A circle with a diameter perpendicular to a chord. | |
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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 
and m 
= m . Since m ∠ AOB = 55°, that would make m = 55° and m 
= 305°. Therefore, m = 27 ½ and m = 152 ½°.
Example 5: Use Figure 8 , in which AB = 8, CD = 8, and OA = 5, to find ON.
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| Figure 8 | A circle with two chords equal in measure. | |
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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.
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