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# 9.6: Segments of Chords

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Find the lengths of chords in a circle.
• Find the measure of arcs in a circle.

## Perpendicular Bisector of a Chord

• The perpendicular bisector of a chord is a diameter.

• The __________________________ of a circle is the perpendicular bisector of any chord.

Proof

Draw two chords, AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} and CD¯¯¯¯¯¯¯¯\begin{align*}\overline{CD}\end{align*} such that AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} is the perpendicular bisector of CD¯¯¯¯¯¯¯¯\begin{align*}\overline{CD}\end{align*}.

The diagram above is a good visual example of this.

We can see that ΔCOEΔDOE\begin{align*}\Delta COE \cong \Delta DOE\end{align*} for any point O\begin{align*}O\end{align*} on chord AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*}.

The congruence of the triangles can be proven by the SAS (Side-Angle-Side) Postulate:

• CE¯¯¯¯¯¯¯¯ED¯¯¯¯¯¯¯¯\begin{align*}\overline{CE} \cong \overline{ED}\end{align*} (it gets bisected by the perpendicular bisector)
• OE¯¯¯¯¯¯¯¯OE¯¯¯¯¯¯¯¯\begin{align*}\overline{OE} \cong \overline{OE}\end{align*} (itself)
• OEC\begin{align*}\angle OEC\end{align*} and OED\begin{align*}\angle OED\end{align*} are right angles (perpendicular makes right angles)

This means that CO¯¯¯¯¯¯¯¯DO¯¯¯¯¯¯¯¯\begin{align*}\overline{CO} \cong \overline{DO}\end{align*} because of CPCTC.

Any point that is equidistant from C\begin{align*}C\end{align*} and D\begin{align*}D\end{align*} lies along AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} by the Perpendicular Bisector Theorem. Since the center of the circle is one such point, it must lie along AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} so AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} is a diameter.

If O\begin{align*}O\end{align*} is the midpoint of AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} then OC¯¯¯¯¯¯¯¯\begin{align*}\overline{OC}\end{align*} and OD¯¯¯¯¯¯¯¯\begin{align*}\overline{OD}\end{align*} are radii of the circle and AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} is a diameter of the circle.

Perpendicular Bisector of a Chord Bisects Intercepted Arc

The perpendicular bisector of a chord bisects the arc intercepted by the chord.

• The _________________________ bisector of a chord also bisects the ________ that is intercepted by that chord.

Proof

We can see that ΔCADΔBAD\begin{align*}\Delta CAD \cong \Delta BAD\end{align*} because of the SAS Postulate:

• BD¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯\begin{align*}\overline{BD} \cong \overline{DC}\end{align*} (it gets bisected by the perpendicular bisector)
• DA¯¯¯¯¯¯¯¯DA¯¯¯¯¯¯¯¯\begin{align*}\overline{DA} \cong \overline{DA}\end{align*} (itself)
• ADB\begin{align*}\angle ADB\end{align*} and ADC\begin{align*}\angle ADC\end{align*} are right angles (perpendicular makes right angles)

This means that DABDAC\begin{align*}\angle DAB \cong \angle DAC\end{align*} because of CPCTC, which means that the intercepted arcs are congruent as well: BEˆCEˆ\begin{align*}\widehat{BE} \cong \widehat{CE}\end{align*}.

The final congruency statement proves that arc BCˆ\begin{align*}\widehat{BC}\end{align*} is bisected, so the proof is complete.

Congruent Chords Equidistant from Center

Congruent chords in the same circle are equidistant (or equal distance) from the center of the circle.

Recall that the definition of distance from a point to a line is the length of the perpendicular segment drawn to the line from the point. CO\begin{align*}CO\end{align*} and DO\begin{align*}DO\end{align*} are these distances, and we must prove that they are equal. Check out the proof on the next page.

Proof

ΔAOEΔBOF\begin{align*}\Delta AOE \cong \Delta BOF\end{align*} because of the SSS Postulate:

• AE¯¯¯¯¯¯¯¯BF¯¯¯¯¯¯¯¯\begin{align*}\overline{AE} \cong \overline{BF}\end{align*} (it is given that these are congruent chords)
• AO¯¯¯¯¯¯¯¯BO¯¯¯¯¯¯¯¯\begin{align*}\overline{AO} \cong \overline{BO}\end{align*} (both are radii of the circle)
• EO¯¯¯¯¯¯¯¯FO¯¯¯¯¯¯¯¯\begin{align*}\overline{EO} \cong \overline{FO}\end{align*} (both are radii of the circle)

Since the triangles are congruent, their corresponding altitudes CO¯¯¯¯¯¯¯¯\begin{align*}\overline{CO}\end{align*} and DO¯¯¯¯¯¯¯¯\begin{align*}\overline{DO}\end{align*} must also be congruent: DO¯¯¯¯¯¯¯¯CO¯¯¯¯¯¯¯¯\begin{align*}\overline{DO} \cong \overline{CO}\end{align*}.

Therefore, AE¯¯¯¯¯¯¯¯BF¯¯¯¯¯¯¯¯\begin{align*}\overline{AE} \cong \overline{BF}\end{align*} and they are equidistant from the center.

Converse of Congruent Chords Theorem

Two chords equidistant from the center of a circle are congruent.

• The converse of what we just proved is also true: if the chords are equidistant from the center, then they are _____________________________.

Example 1

CE=12\begin{align*}CE = 12\end{align*} inches, and is 3 inches from the center of circle O\begin{align*}O\end{align*}.

Find the radius of the circle.

Draw the radius OC¯¯¯¯¯¯¯¯\begin{align*}\overline{OC}\end{align*}.

OC¯¯¯¯¯¯¯¯\begin{align*}\overline{OC}\end{align*} is the hypotenuse of the right triangle \begin{align*}\Delta COT\end{align*}.

\begin{align*}OT =\end{align*} _________ inches and \begin{align*}CT =\end{align*} half of 12 inches \begin{align*}=\end{align*} _________ inches.

Apply the Pythagorean Theorem:

\begin{align*}(OC)^2 & = (OT)^2 + (CT)^2\\ (OC)^2 & = 3^2 + 6^2 = 9 + 36 = 45\\ OC & = \sqrt{45} = 3 \sqrt{5} \approx 6.7 \ inches\end{align*}

The radius of circle \begin{align*}O\end{align*} is about 6.7 inches long.

Example 2

Two concentric circles have radii of 6 inches and 10 inches. A segment tangent to the smaller circle is a chord of the larger circle. What is the length of the segment?

Start by drawing a figure that represents the problem like the one above.

\begin{align*}OC =\end{align*} __________ inches and \begin{align*}OB =\end{align*} __________ inches

\begin{align*}\Delta COB\end{align*} is a right triangle because the radius \begin{align*}\overline{OC}\end{align*} of the smaller circle is perpendicular to the tangent \begin{align*}\overline{AB}\end{align*} at point \begin{align*}C\end{align*}.

Apply the Pythagorean Theorem:

\begin{align*}(OC)^2 + (BC)^2 & = (OB)^2\\ 6^2 + (BC)^2 & = 10^2\\ 36 + (BC)^2 & = 100\\ (BC)^2 & = 100 - 36 = 64\\ BC & = \sqrt{64} = 8 \ inches\end{align*}

You learned earlier in this lesson that because \begin{align*}\overline{OC}\end{align*} is a radius (which is part of a diameter), it is also a perpendicular bisector of the chord \begin{align*}\overline{AB}\end{align*}, which means that \begin{align*}BC\end{align*} is half of \begin{align*}AB\end{align*}, or \begin{align*}AB = 2BC\end{align*}.

Therefore, \begin{align*}AB = 2(8) = 16\end{align*} inches.

1. True or false: The diameter of a circle is perpendicular to every chord in the circle.

2. True or false: The perpendicular bisector of a chord also bisects the arc that is intercepted by the chord.

3. True or false: Two chords that are the same distance away from the center of a circle are also the same length.

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8 , 9 , 10
Date Created:
Jan 13, 2015
Jan 13, 2015
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