15.4 Segment Relationships in Circles

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Name
Class
Date
15.4 Segment Relationships
in Circles
Essential Question: What are the relationships between the segments in circles?
Resource
Locker
Exploring Segment Length Relationships
in Circles
Explore
Any segment connecting two points on a circle is a chord. In some cases, two chords drawn inside
the same circle will intersect, creating four segments. In the following activity, you will look for a
pattern in how these segments are related and form a conjecture.
A
Using geometry
software or a compass and straightedge, construct circle A with two
_
_
chords CD and EF that intersect inside the circle. Label the intersection point G.
F
G
D
Circle 1
―――
C
A
Students’ circles may vary.
B
E
© Houghton Mifflin Harcourt Publishing Company
B
Repeat your construction with two more circles. Vary the size of the circles
and where you put the intersecting chords inside them.
Circle 2
―――
Circle 3
―――
Module 15
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C
DO NOT
Correcti
Fill in the chart with the lengths of the segments measured to the nearest millimeter and
calculate their products.
Students’ results may vary. Possible results are shown.

DG ⋅ GC
EG ⋅ GF
16.0
449
449
14.1
494
494
DG
GC
EG
GF
Circle 1
13.5
26.8
25.2
14.4
Circle 2
15.0
29.9
28.1
Circle 3
19.0
26.0
35.0
362
362
Look for a pattern among the measurements and calculations of the segments. From the
table, it appears that DG ⋅ GC will always equal EG ⋅ GF .
Reflect
1.
Discussion Compare your results with those of your classmates. What do you notice?
Possible answer: Everyone answered Step D the same; the values of DG ⋅ GC and EG ⋅ GF
were always the same.
2.
What conjecture can you make about the products of the segments of two chords that
intersect inside a circle?
The products of the lengths of the segments of two chords that intersect inside
Conjecture:
a circle will be equal.
Explain 1
Applying the Chord-Chord Product Theorem
In the Explore, you discovered a pattern in the relationship between the parts of two chords
that intersect inside a circle. In this Example, you will apply the following theorem to
solve problems.
If two chords intersect inside a circle, then the products
of the lengths of the segments of the chords are equal.
B
C
E
D
AE · EB = CE · ED
A
Module 15
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Chord-Chord Product Theorem
Lesson 4
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Example 1 Find the value of

x and the length of each chord.
C
Set up an equation according to the Chord-Chord Product Theorem and
solve for x.
6
B
3
CE ⋅ ED = AE ⋅ EB
x
6(2) = 3(x)
E
2
A
12 = 3x
D
4=x
Add the segment lengths to find the length of each chord.
CD = CE + ED = 6 + 2 = 8
AB = AE + EB = 4 + 3 = 7

H
Set up an equation according to the Chord-Chord Product Theorem and
solve for x:
HG ⋅ GJ = KG ⋅ GI
9
( 8 )=
6
(x)
I
9
G
6
K
x
8
J
72 = 6x
12 = x
Add the segment lengths together to find the lengths of each chord:
HJ = HG + GJ =
9 + 8=
KI = KG + GI = 6 +
17
12 = 18
Your Turn
3.
Given AD = 12. Find the value of x and the length of each chord.
AE ⋅ ED = CE ⋅ CB
B
A
3(12 - 3) = 14(x)
4
3
E
x
27 = 14x
C
1.93 ≈ x
D
CB ≈ 15.93
AD = 12 (given)
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Explain 2
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Proving the Secant-Secant Product Theorem
A secant is any line that intersects a circle at exactly two points. A secant segment is part of a secant
line with at least one point on the circle. A secant segment that lies in the exterior of the circle with
one point on the circle is called an external secant segment. Secant segments drawn from the same
point in the exterior of a circle maintain a certain relationship that can be stated as a theorem.
Secant-Secant Product Theorem
If two secants intersect in the exterior of a circle, then the
product of the lengths of one secant segment and its
external segment equals the product of the lengths of the
other secant segment and its external segment.
A
B
C
E
D
AE · BE = CE · DE
Example 2 Use similar triangles to prove the Secant-Secant Product Theorem.
Step 1 Identify the segments in the diagram. The whole secant segments in this
¯
¯
AE
diagram are
and CE .
¯
BE
and
The external secant segments in this diagram are
¯
DE
.
Step 2
Given the diagram as shown, prove that
Prove: AE ⋅ BE = CE ⋅ DE
A
B
C
AE ⋅ BE = CE ⋅ DE.
¯ and CB
¯. ∠EAD and ∠ECB both
AD
Proof: Draw auxiliary line segments
⁀
BD , so ∠ EAD ≅ ∠ ECB . ∠E ≅ ∠E by the
intercept
Reflexive
Property. Thus, △EAD ∼ △ECB by
E
D
A
B
C
E
D
the AA Triangle Similarity Theorem . Therefore, corresponding sides
DE
Multiplication Property of Equality,
AE = _
_
. By the
BE
CE
AE
DE
BE(CE) ⋅ ___
= ___
⋅ BE(CE), and thus AE ⋅ BE = CE ⋅ DE.
BE
CE
Reflect
4.
Rewrite the Secant-Secant Theorem in your own words. Use a diagram or shortcut
notation to help you remember what it means.
Possible answer: For two secant segments drawn to a circle from the same point, (whole)
(outside) of first segment equals (whole)(outside) of second segment.
5.
are proportional, so
Discussion: Suppose that two secants are drawn so that they intersect on the circle.
Can you determine anything about the lengths of the segments formed? Explain.
No; if the endpoints of the secants lie outside the circle, the lengths outside the circle can
be extended without changing the segments inside the circle. So no relationship exists
using the external segments and whole secant segments.
Module15
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Lesson4
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Applying the Secant-Secant Product Theorem
Explain 3
You can use the Secant-Secant Product Theorem to find unknown measures of secants and
secant segments by setting up an equation.

x and the length of each secant segment.
Set up an equation according to the Secant-Secant Product Theorem and
solve for x.
C
x
AC ⋅ AB = AE ⋅ AD
5 B
A
(5 + x)(5) = (12)(6)
6
6
D
E
5x + 25 = 72
5x = 47
x = 9.4
Add the segments together to find the lengths of each secant segment.
AC = 5 + 9.4 = 14.4; AE = 6 + 6 = 12

Set up an equation according to the Secant-Secant Product Theorem and
solve for x.
UP ⋅ TP = SP ⋅ RP
( x + 7 )(7) = (
14
S
)(6)
6
R
8
x
P
7
T
U
7 x + 49 = 84
7 x = 35
x= 5
Add the segments together to find the lengths of each secant segment.
UP = 7 + 5 = 12 ; SP = 8 + 6 = 14
Your Turn
Find the value of x and the length of each secant segment.
6.
P
5
4
Q
5.4
R
7.
H
N
S
9
T
4
J
6
3.5 P 3.5 M 5
NL ⋅ ML = HL ⋅ JL
PT ⋅ PS = PR ⋅ PQ
(4 + x)(4) = (10.4)(5)
(5+2x)(5) =(10)(6)
PT = 13; PR = 10.4
HL = 10; NL = 12
x = 3.5
x=9
Module15
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Applying the Secant-Tangent Product Theorem
Explain 4
A similar theorem applies when both a secant segment and tangent segment are drawn to
a circle from the same exterior point. A tangent segment is a segment of a tangent line
with exactly one endpoint on the circle.
Secant-Tangent Product Theorem
If a secant and a tangent intersect in the exterior of a circle, then
the product of the lengths of the secant segment and its external
segment equals the length of the tangent segment squared.
A
B
C
AC · BC = DC 2
D
x.
 Given the diameter of the Earth as 8,000 miles, a satellite’s
orbit
_is 6,400 miles above the Earth. Its range, shown
by SP, is a tangent segment.
Set up an equation according to the Secant-Tangent
Product Theorem and solve for x:
SA ⋅ SE = SP
(8000 + 6400)(6400) = x 2
(14400)(6400) = x
S
2
E
92,160,000 = x 2
Since distance must be positive, the value of
P
8000
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
©cristimatei/iStockPhoto.com
±9600 = x
x must be 9600 miles.
A
 Set up an equation according to the Secant-Tangent Product Theorem and solve for x:
BD ⋅ BC = BA 2
( x + 2 )(2) =5
2
A
2 x + 4 = 25
5
2 x = 21
B
2
C
10.5
D
x = 10.5
Module15
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Lesson4
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Reflect
8.
Compare and contrast the Secant-Secant Product Theorem with the
Secant-Tangent Product Theorem.
Both theorems involve equations where two lengths are being multiplied together on
each side of the equal side. In the Secant-Tangent Product Theorem, since the tangent
segment has only one length to consider, it is multiplied by itself.
Your Turn
Find the value of x.
9.
On a bird-watching trip, you travel along a path tangent to a circular pond
to a lookout station that faces a hawk’s nest. Given the measurements in the
diagram on your bird-watching map, how far is the nest from the lookout
station?
x
25yd
Lookout Station
Nest
100yd
(x)(25) = 100 2
25x = 10000
x = 400
The nest is 400 yards from the lookout station.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Roy Toft/
National Geographic/Getty Images
Path
10.
A
6
B
C
2 10 ≈ 6.32
D
Module 15
4
AC · BC = CD 2
(10)(4) = x 2
40 = x 2
―
±2 √10 = x
―
Since distance must be positive,
the value of x must be 2 √10 ≈ 6.32.
821
Lesson 4
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DO NOT
Correcti
Elaborate
11. How is solving for y in the following diagram different from the Example 3?
G
H
6
7.5
J
10
F
For the secant segment in this case, both the secant external segment and the whole
segment involve the number x, so the equation would be GJ ⋅ HJ = FJ 2, or (6 + y)y = 10 2.
Then, we would multiply y by (6 + y) by y would result in a quadratic equation to solve.
12. A circle is constructed with two secant segments that intersect outside the circle. If both external secant
segments are equal, is it reasonable to conclude that both secant segments are equal? Explain.
Yes. Suppose the length of the external part of each secant segment equals a and the
lengths of the whole secant segments are b and c. By the Secant-Secant Product Theorem,
b(a) = c(a). Dividing both sides by a leaves b = c, which means that the two secant
segments must be equal.
13. Essential Question Check-In How are the theorems in this lesson related?
All three theorems involve segments drawn intersecting with a circle. They also involve
equations where you multiply two segment lengths on each side of an equal sign. The two
segments either intersect inside or outside the circle.
Module 15
822
Lesson 4
23/03/14 3:07 AM

15.4 Segment Relationships in Circles

Transcription

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