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9.3: Properties of Chords

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Learning Objectives

  • Find the lengths of chords in a circle.
  • Discover properties of chords and arcs.

Review Queue

  1. Draw a chord in a circle.
  2. Draw a diameter in the circle from #1. Is a diameter a chord?
  3. \triangle ABC is an equilateral triangle in \bigodot A. Find m \widehat{BC} and m \widehat{BDC}.
  4. \triangle ABC and \triangle ADE are equilateral triangles in \bigodot A. List a pair of congruent arcs and chords.

Know What? To the right is the Gran Teatro Falla, in Cadiz, Andalucía, Spain. 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, A-E. \bigodot A \cong \bigodot E and \bigodot B \cong \bigodot C \cong \bigodot D. Each window is topped with a 240^\circ 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?

Recall from the first section, that 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.

Congruent Chords & Congruent Arcs

From #4 in the Review Queue above, we noticed that \overline{BC} \cong \overline{DE} and \widehat{BC} \cong \widehat{DE}. This leads to our first theorem.

Theorem 10-3: 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 Theorem 10-3 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, \overline{BE} \cong \overline{CD} and \widehat{BE} \cong \widehat{CD}. In the second picture, we have \triangle BAE \cong \triangle CAD because the central angles are congruent and \overline{BA} \cong \overline{AC} \cong \overline{AD} \cong \overline{AE} because they are all radii (SAS). By CPCTC, \overline{BE} \cong \overline{CD}.

Example 1: Use \bigodot A to answer the following.

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

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

Solution:

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

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

Investigation 9-2: Perpendicular Bisector of a Chord

Tools Needed: paper, pencil, compass, ruler

  1. Draw a circle. Label the center A.
  2. Draw a chord in \bigodot A. Label it \overline{BC}.
  3. Find the midpoint of \overline{BC} by using a ruler. Label it D.
  4. Connect A and D to form a diameter. How does \overline{AD} relate to the chord, \overline{BC}?

Theorem 10-4: The perpendicular bisector of a chord is also a diameter.

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

Theorem 10-5: If a diameter is perpendicular to a chord, then the diameter bisects the chord and its corresponding arc.

Example 2: Find the value of x and y.

Solution: The diameter here is also perpendicular to the chord. From Theorem 10-5, x = 6 and y = 75^\circ.

Example 3: Is the converse of Theorem 10-4 true?

Solution: The converse of Theorem 10-4 would be: A diameter is also the perpendicular bisector of a chord. This is not a true statement, see the counterexample to the right.

Example 4: Algebra Connection Find the value of x and y.

Solution: Because the diameter is perpendicular to the chord, it also bisects the chord and the arc. Set up an equation for x and y.

(3x-4)^\circ &= (5x-18)^\circ && \ y+4=2y+1\\14^\circ &= 2x && \qquad 3=y\\7^\circ &= x

Equidistant Congruent Chords

Investigation 9-3: Properties of Congruent Chords

Tools Needed: pencil, paper, compass, ruler

  1. 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.
  2. From the center, draw the perpendicular segment to \overline{AB} and \overline{CD}. You can either use your ruler, a protractor or Investigation 3-2 (Constructing a Perpendicular Line through a Point not on the line. We will show arc marks for Investigation 3-2.
  3. Erase the arc marks and lines beyond the points of intersection, leaving \overline{FE} and \overline{EG}. Find the measure of these segments. What do you notice?

Theorem 10-6: 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 FE = EG means that \overline{AB} and \overline{CD} are equidistant to the center and \overline{AB} \cong \overline{CD}.

Example 5: Algebra Connection Find the value of x.

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

6x-7 &= 35\\6x &= 42\\x &= 7

Example 6: BD = 12 and AC = 3 in \bigodot A. Find the radius and m \widehat{BD}.

Solution: First find the radius. In the picture, \overline{AB} is a radius, so we can use the right triangle \triangle ABC, such that \overline{AB} is the hypotenuse. From 10-5, BC = 6.

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

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

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

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

Know What? Revisited In the picture, the chords from \bigodot A and \bigodot E are congruent and the chords from \bigodot B, \bigodot C, and \bigodot D are also congruent. We know this from Theorem 10-3. All five chords are not congruent because all five circles are not congruent, even though the central angle for the circles is the same.

Review Questions

  1. Two chords in a circle are perpendicular and congruent. Does one of them have to be a diameter? Why or why not? Fill in the blanks.
  2. \underline{\;\;\;\;\;\;\;} \cong \overline{DF}
  3. \widehat{AC} \cong \underline{\;\;\;\;\;\;\;}
  4. \widehat{DJ} \cong \underline{\;\;\;\;\;\;\;}
  5. \underline{\;\;\;\;\;\;\;} \cong \overline{EJ}
  6. \angle AGH \cong \underline{\;\;\;\;\;\;\;}
  7. \angle DGF \cong \underline{\;\;\;\;\;\;\;}
  8. List all the congruent radii in \bigodot G.

Find the value of the indicated arc in \bigodot A.

  1. m \widehat{BC}
  2. m \widehat{BD}
  3. m \widehat{BC}
  4. m \widehat{BD}
  5. m \widehat{BD}
  6. m \widehat{BD}

Algebra Connection Find the value of x and/or y.

  1. AB = 32
  2. Find m \widehat{AB} in Question 18. Round your answer to the nearest tenth of a degree.
  3. Find m \widehat{AB} in Question 23. Round your answer to the nearest tenth of a degree.

In problems 26-28, what can you conclude about the picture? State a theorem that justifies your answer. You may assume that A is the center of the circle.

  1. Trace the arc below onto your paper then follow the steps to locate the center using a compass and straightedge.
    1. Use your straightedge to make a chord in the arc.
    2. Use your compass and straightedge to construct the perpendicular bisector of this chord.
    3. Repeat steps a and b so that you have two chords and their perpendicular bisectors.
    4. What is the significance of the point where the perpendicular bisectors intersect?
    5. Verify your answer to part d by using the point and your compass to draw the rest of the circle.
  2. Algebra Connection Let’s repeat what we did in problem 29 using coordinate geometry skills. Given the points A(-3, 5), B(5, 5) and C(4, -2) on the circle (an arc could be drawn through these points from A to C). The following steps will walk you through the process to find the equation of the perpendicular bisector of a chord, and use two of these perpendicular bisectors to locate the center of the circle. Let’s first find the perpendicular bisector of chord \overline{AB}.
    1. Since the perpendicular bisector passes through the midpoint of a segment we must first find the midpoint between A and B.
    2. Now the perpendicular line must have a slope that is the opposite reciprocal of the slope of \overleftrightarrow{AB}. Find the slope of \overleftrightarrow{AB} and then its opposite reciprocal.
    3. Finally, you can write the equation of the perpendicular bisector of \overline{AB} using the point you found in part a and the slope you found in part b.
    4. Repeat steps a-c for chord \overline{BC}.
    5. Now that we have the two perpendicular bisectors of the chord we can use algebra to find their intersection. Solve the system of linear equations to find the center of the circle.
    6. Find the radius of the circle by finding the distance from the center (point found in part e) to any of the three given points on the circle.
  3. Find the measure of \widehat{AB} in each diagram below.

Review Queue Answers

1 & 2. Answers will vary

3. m\widehat{BC}=60^\circ, m\widehat{BDC}=300^\circ

4. \overline{BC} \cong \overline{DE} and \widehat{BC} \cong \widehat{DE}

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Date Created:

Feb 23, 2012

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Sep 22, 2014
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