9.6: Segments of Chords
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, \begin{align*}\overline{AB}\end{align*}
The diagram above is a good visual example of this.
We can see that \begin{align*}\Delta COE \cong \Delta DOE\end{align*}
The congruence of the triangles can be proven by the SAS (SideAngleSide) Postulate:

\begin{align*}\overline{CE} \cong \overline{ED}\end{align*}
CE¯¯¯¯¯¯¯¯≅ED¯¯¯¯¯¯¯¯ (it gets bisected by the perpendicular bisector) 
\begin{align*}\overline{OE} \cong \overline{OE}\end{align*}
OE¯¯¯¯¯¯¯¯≅OE¯¯¯¯¯¯¯¯ (itself) 
\begin{align*}\angle OEC\end{align*}
∠OEC and \begin{align*}\angle OED\end{align*}∠OED are right angles (perpendicular makes right angles)
This means that \begin{align*}\overline{CO} \cong \overline{DO}\end{align*}
Any point that is equidistant from \begin{align*}C\end{align*}
If \begin{align*}O\end{align*}
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 \begin{align*}\Delta CAD \cong \Delta BAD\end{align*}

\begin{align*}\overline{BD} \cong \overline{DC}\end{align*}
BD¯¯¯¯¯¯¯¯≅DC¯¯¯¯¯¯¯¯ (it gets bisected by the perpendicular bisector) 
\begin{align*}\overline{DA} \cong \overline{DA}\end{align*}
DA¯¯¯¯¯¯¯¯≅DA¯¯¯¯¯¯¯¯ (itself) 
\begin{align*}\angle ADB\end{align*}
∠ADB and \begin{align*}\angle ADC\end{align*}∠ADC are right angles (perpendicular makes right angles)
This means that \begin{align*}\angle DAB \cong \angle DAC\end{align*}
The final congruency statement proves that arc \begin{align*}\widehat{BC}\end{align*}
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. \begin{align*}CO\end{align*}
Proof
\begin{align*}\Delta AOE \cong \Delta BOF\end{align*}

\begin{align*}\overline{AE} \cong \overline{BF}\end{align*}
AE¯¯¯¯¯¯¯¯≅BF¯¯¯¯¯¯¯¯ (it is given that these are congruent chords) 
\begin{align*}\overline{AO} \cong \overline{BO}\end{align*}
AO¯¯¯¯¯¯¯¯≅BO¯¯¯¯¯¯¯¯ (both are radii of the circle) 
\begin{align*}\overline{EO} \cong \overline{FO}\end{align*}
EO¯¯¯¯¯¯¯¯≅FO¯¯¯¯¯¯¯¯ (both are radii of the circle)
Since the triangles are congruent, their corresponding altitudes \begin{align*}\overline{CO}\end{align*}
Therefore, \begin{align*}\overline{AE} \cong \overline{BF}\end{align*}
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
\begin{align*}CE = 12\end{align*}
Find the radius of the circle.
Draw the radius \begin{align*}\overline{OC}\end{align*}
\begin{align*}\overline{OC}\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.
Reading Check:
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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