# 9.4: Chords in Circles

**At Grade**Created by: CK-12

**Practice**Chords in Circles

What if you were asked to geometrically consider the Gran Teatro Falla, in Cadiz, Andalucía, Spain, pictured below? This theater was built in 1905 and hosts several plays and concerts. It is an excellent example of circles in architecture. Notice the five windows, \begin{align*}A-E\end{align*}. \begin{align*}\bigodot A \cong \bigodot E\end{align*} and \begin{align*}\bigodot B \cong \bigodot C \cong \bigodot D\end{align*}. Each window is topped with a \begin{align*}240^\circ\end{align*} arc. The gold chord in each circle connects the rectangular portion of the window to the circle. Which chords are congruent? How do you know? After completing this Concept, you'll be able to use properties of chords to answer questions like these.

### Watch This

CK-12 Foundation: Chapter9ChordsinCirclesA

Brightstorm: Chords and a Circle's Center

### Guidance

A **chord** is a line segment whose endpoints are on a circle. A diameter is the longest chord in a circle. There are several theorems that explore the properties of chords.

**Chord Theorem #1:** In the same circle or congruent circles, minor arcs are congruent if and only if their corresponding chords are congruent.

Notice the “if and only if” in the middle of the theorem. This means that Chord Theorem #1 is a biconditional statement. Taking this theorem one step further, any time two central angles are congruent, the chords and arcs from the endpoints of the sides of the central angles are also congruent. In both of these pictures, \begin{align*}\overline{BE} \cong \overline{CD}\end{align*} and \begin{align*}\widehat{BE} \cong \widehat{CD}\end{align*}. In the second picture, we have \begin{align*}\triangle BAE \cong \triangle CAD\end{align*} because the central angles are congruent and \begin{align*}\overline{BA} \cong \overline{AC} \cong \overline{AD} \cong \overline{AE}\end{align*} because they are all radii (SAS). By CPCTC, \begin{align*}\overline{BE} \cong \overline{CD}\end{align*}.

##### Investigation: Perpendicular Bisector of a Chord

Tools Needed: paper, pencil, compass, ruler

- Draw a circle. Label the center \begin{align*}A\end{align*}.
- Draw a chord in \begin{align*}\bigodot A\end{align*}. Label it \begin{align*}\overline{BC}\end{align*}.
- Find the midpoint of \begin{align*}\overline{BC}\end{align*} by using a ruler. Label it \begin{align*}D\end{align*}.
- Connect \begin{align*}A\end{align*} and \begin{align*}D\end{align*} to form a diameter. How does \begin{align*}\overline{AD}\end{align*} relate to the chord, \begin{align*}\overline{BC}\end{align*}?

**Chord Theorem #2:** The perpendicular bisector of a chord is also a diameter.

In the picture to the left, \begin{align*}\overline{AD} \bot \overline{BC}\end{align*} and \begin{align*}\overline{BD} \cong \overline{DC}\end{align*}. From this theorem, we also notice that \begin{align*}\overline{AD}\end{align*} also bisects the corresponding arc at \begin{align*}E\end{align*}, so \begin{align*}\widehat{BE} \cong \widehat{EC}\end{align*}.

**Chord Theorem #3:** If a diameter is perpendicular to a chord, then the diameter bisects the chord and its corresponding arc.

##### Investigation: Properties of Congruent Chords

Tools Needed: pencil, paper, compass, ruler

- Draw a circle with a radius of 2 inches and two chords that are both 3 inches. Label as in the picture to the right.
*This diagram is drawn to scale.* - From the center, draw the perpendicular segment to \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{CD}\end{align*}.

- Erase the arc marks and lines beyond the points of intersection, leaving \begin{align*}\overline{FE}\end{align*} and \begin{align*}\overline{EG}\end{align*}. Find the measure of these segments. What do you notice?

**Chord Theorem #4:** In the same circle or congruent circles, two chords are congruent if and only if they are equidistant from the center.

Recall that two lines are equidistant from the same point if and only if the shortest distance from the point to the line is congruent. The shortest distance from any point to a line is the perpendicular line between them. In this theorem, the fact that \begin{align*}FE = EG\end{align*} means that \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{CD}\end{align*} are equidistant to the center and \begin{align*}\overline{AB} \cong \overline{CD}\end{align*}.

#### Example A

Use \begin{align*}\bigodot A\end{align*} to answer the following.

a) If \begin{align*}m \widehat{BD}= 125^\circ\end{align*}, find \begin{align*}m \widehat{CD}\end{align*}.

b) If \begin{align*}m \widehat{BC}= 80^\circ\end{align*}, find \begin{align*}m \widehat{CD}\end{align*}.

**Solutions:**

a) From the picture, we know \begin{align*}BD = CD\end{align*}. Because the chords are equal, the arcs are too. \begin{align*}m \widehat{CD}= 125^\circ\end{align*}.

b) To find \begin{align*}m \widehat{CD}\end{align*}, subtract \begin{align*}80^\circ\end{align*} from \begin{align*}360^\circ\end{align*} and divide by 2. \begin{align*}m \widehat{CD}=\frac{360^\circ - 80^\circ}{2}=\frac{280^\circ}{2}=140^\circ\end{align*}

#### Example B

Find the value of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

The diameter here is also perpendicular to the chord. From Chord Theorem #3, \begin{align*}x = 6\end{align*} and \begin{align*}y = 75^\circ\end{align*}.

#### Example C

Find the value of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

Because the diameter is perpendicular to the chord, it also bisects the chord and the arc. Set up an equation for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

\begin{align*}(3x-4)^\circ &= (5x-18)^\circ && \ y+4=2y+1\\ 14^\circ &= 2x && \qquad 3=y\\ 7^\circ &= x\end{align*}

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter9ChordsinCirclesB

#### Concept Problem Revisited

In the picture, the chords from \begin{align*}\bigodot A\end{align*} and \begin{align*}\bigodot E\end{align*} are congruent and the chords from \begin{align*}\bigodot B, \bigodot C\end{align*}, and \begin{align*}\bigodot D\end{align*} are also congruent. We know this from Chord Theorem #1. All five chords are not congruent because all five circles are not congruent, even though the central angle for the circles is the same.

### Vocabulary

A ** circle** is the set of all points that are the same distance away from a specific point, called the

**. A**

*center***is the distance from the center to the outer rim of a circle. A**

*radius***is a line segment whose endpoints are on a circle. A**

*chord***is a chord that passes through the center of the circle.**

*diameter*### Guided Practice

1. Is the converse of Chord Theorem #2 true?

2. Find the value of \begin{align*}x\end{align*}.

3. \begin{align*}BD = 12\end{align*} and \begin{align*}AC = 3\end{align*} in \begin{align*}\bigodot A\end{align*}. Find the radius and \begin{align*}m \widehat{BD}\end{align*}.

**Answers:**

1. The converse of Chord Theorem #2 would be: A diameter is also the perpendicular bisector of a chord. This is not a true statement, see the counterexample to the right.

2. Because the distance from the center to the chords is congruent and perpendicular to the chords, then the chords are equal.

\begin{align*}6x-7 &= 35\\ 6x &= 42\\ x &= 7\end{align*}

3. First find the radius. In the picture, \begin{align*}\overline{AB}\end{align*} is a radius, so we can use the right triangle \begin{align*}\triangle ABC\end{align*}, such that \begin{align*}\overline{AB}\end{align*} is the hypotenuse. From Chord Theorem #3, \begin{align*}BC = 6\end{align*}.

\begin{align*}3^2+6^2 &= AB^2\\ 9+36 &= AB^2\\ AB &= \sqrt{45}=3 \sqrt{5}\end{align*}

In order to find \begin{align*}m \widehat{BD}\end{align*}, we need the corresponding central angle, \begin{align*}\angle BAD\end{align*}. We can find half of \begin{align*}\angle BAD\end{align*} because it is an acute angle in \begin{align*}\triangle ABC\end{align*}. Then, multiply the measure by 2 for \begin{align*}m \widehat{BD}\end{align*}.

\begin{align*}\tan^{-1} \left( \frac{6}{3} \right) &= m \angle BAC\\ m \angle BAC & \approx 63.43^\circ\end{align*}

This means that \begin{align*}m \angle BAD \approx 126.9^\circ\end{align*} and \begin{align*}m \widehat{BD} \approx 126.9^\circ\end{align*} as well.

### Practice

Find the value of the indicated arc in \begin{align*}\bigodot A\end{align*}.

- \begin{align*}m \widehat{BC}\end{align*}
- \begin{align*}m \widehat{BD}\end{align*}
- \begin{align*}m \widehat{BC}\end{align*}
- \begin{align*}m \widehat{BD}\end{align*}
- \begin{align*}m \widehat{BD}\end{align*}
- \begin{align*}m \widehat{BD}\end{align*}

** Algebra Connection** Find the value of \begin{align*}x\end{align*} and/or \begin{align*}y\end{align*}.

- \begin{align*}AB = 32\end{align*}
- Find \begin{align*}m \widehat{AB}\end{align*} in Question 10. Round your answer to the nearest tenth of a degree.
- Find \begin{align*}m \widehat{AB}\end{align*} in Question 15. Round your answer to the nearest tenth of a degree.

In problems 18-20, what can you conclude about the picture? State a theorem that justifies your answer. You may assume that \begin{align*}A\end{align*} is the center of the circle.

- Find the measure of \begin{align*}\widehat{AB}\end{align*} in each diagram below.

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### Image Attributions

Here you'll learn theorems about chords in circles and how to apply them.