Transcript Slide 1

Geometry
Bisector of
Triangles
CONFIDENTIAL
1
Warm up
Determine whether each point is on the perpendicular
bisector of the segment with endpoints S(0,8) and T(4,0).
1) X(0,3)
2) Y(-4,1)
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3) Z(-8,-2)
2
Bisector of Triangles
Since a triangle has three sides, it has
three perpendicular bisector. When you
construct the perpendicular bisectors, you
find that they have an interesting property.
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1
2
3
B
B
P
C
A
C
Draw a large
scalene acute
triangle ABC on a
piece of patty
paper.
A
Fold the
perpendicular
bisector of each
side.
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C
Label the point
where the three
perpendicular
bisectors
intersect as P.
4
When three or more lines intersect at one point,
the lines are said to be concurrent. The point of
concurrency is the point where they intersect.
In the construction, you saw that the three
perpendicular bisectors of a triangle are
concurrent. This point of concurrency is the
circumcenter of the triangle.
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Circumcenter Theorem
Theorem 2.1
B
The circumcenter of a triangle is
equidistant from the vertices of the
triangle.
P
PA = PB = PC
A
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C
6
The circumcenter can be inside the triangle, Out the triangle,
or on the triangle.
P
P
P
Acute triangle
Obtuse triangle
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Right triangle
7
The circumcenter of ∆ABC is the center of its
circumscribed circle. A circle that contains all the vertices
of a polygon is circumscribed about the polygon.
B
P
A
C
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Circumcenter Theorem
A
l
n
Given: Line l , m, and n are the
perpendicular bisec tor of AB, BC,
and AC, respectively.
Prove: PA = PB = PC
B
Proof:
P is the circumcenter of ABC. Since P
lies on the perpendicular bisector of AB,
PA = PB by the Perpendicular Bisector
Theorem. Similarly, P also lies on t he
perpendicular bisector of BC, so PB = PC.
Therefore PA = PB = PC by the
Transit ive Propert y of Equality.
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P
C
m
9
Using Properties of
Perpendicular Bisector
H
KZ, LZ, and MZ are the perpendicular bisectors
of GHJ. Find HZ.
18.6
Z is the circumcenter of ∆GHJ. By the
Circumcenter Theorem, Z is equidistant from
the vertices of ∆GHJ.
K
HZ = GZ
Circumcenter Thm.
HZ = 19.9
Substitute 19.9 for GZ. 19.9
G
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Z
L
9.5
M
J
14.5
10
Now you try!
1)Use the diagram. Find each length.
a) GM
b) GK
c) JZ
H
18.6
K
Z
L
9.5
19.9
G
M 14.5
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J
11
Finding the Circumcenter of a Triangle.
Finding the circumcenter of ∆RSO with vertices
R(-6,0), S(0,4), and O (0,0).
Step 1 Graph the triangle.
y
X = -4
Y=2
S
(-3,2)
R 4
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O
x
Next page:
12
Step 2 Find equation for two perpendicular bisectors. Since
Two sides of the triangle lie along the axes, use the graph to
find the perpendicular bisectors of these two sides. The
perpendicular bisector of RO is x = -3, and the perpendicular
bisector of OS is y =2.
y
X = -4
Y=2
S
(-3,2)
R 4
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O
x
Next page:
13
Step 3 Find the intersection of the two equations.
The lines x = -3 and y = z intersect at (-3,2), the
circumcenter of ∆RSO.
y
X = -4
Y=2
S
(-3,2)
R 4
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O
x
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Now you try!
2) Find the circumcenter of ∆GOH with vertices
G(0,-9), O(0,0), and H(8,0).
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Theorem 2.2
Incenter Theorem
The incenter of a triangle is equidistant from the
sides of the triangle.
B
PX = PY = PZ
Z
Y
P
A
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X
C
16
Unlike the circumcenter, the incenter is always
inside the triangle.
P
P
Acute triangle
Obtuse triangle
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P
Right triangle
17
The incenter is the center of the triangle’s inscribed
circle. A circle inscribed in a polygon intersects each line
that contains a side of the polygon at exactly one point.
B
P
A
C
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Using Properties of Angle
Bisectors
K
A) The distance from V to KL
7.3
V is the incenter of ∆JKL. By the
Incenter Theorem, V is equidistant from
the sides of ∆JKL.
W
The distance from V to JK is 7.3.
V
106˚
So the distance from V to KL is also 7.3
J
19˚
L
Next page:
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B)
m/ Vkl
JV is the bisector of m/ kJL.
Substitute 19˚ for m / VJL. ∆
m/ kJL = 2(19˚) = 38˚
sum Thm.
m/ kJL + m/ JLK + m/ JKL = 180˚ Substitute the
7.3
given values.
38 + 106 + m/ JKL = 180˚
Subtract 144˚ from W
both sides.
m/ JKL = 36˚
K
m/ kJL = 2m / VJL
m/ Vkl = ½ m/ JKL
KV is the bisector of m/ JKL
m/ Vkl = ½ (36˚) = 18˚
V
106˚
Substitute 36˚ for
m/ JKL.
19˚
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L
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Now you try!
3) QX and RX are angle bisectors ∆PQR. Find each
measure.
a) The distance from X to PQ
b) m/ PQX
Q
X
R
P
Y
19.2
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12˚
21
Community Application
The city of Odessa will host a fireworks display for the next
Fourth of July celebration. Draw a sketch to show where the
display should be positioned so that it is the same distance
from all three viewing location A, B, and C on the map. Justify
your sketch.
A
Let the three viewing locations
be vertices of a triangle. By
the Circumcenter Theorem, the
circumcenter of the triangle is
equidistant from the vertices.
C
B
Next page:
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Trace the map. Draw the triangle formed by the
viewing locations. To find the circumcenter, find
the perpendicular bisectors of each side. The
position of the display is the circumcenter, F.
A
C
F
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B
23
Now you try!
4) A City plans to build a firefighters’ monument in the
park between three streets. Draw a sketch to show where
the city should place the monument so that it is the same
distance from all three streets. Justify your sketch.
King Boulevard
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Now some practice problems for
you!
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Assessment
SN, TN, and VN are the perpendicular bisectors of
∆PQR. Find each length.
Q
1) NR
2) RV
3) TR
4) QN
3.95
S
N
5.64
P
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T
4.03
5.47
V
R
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Find the circumcenter of a triangle with the
given vertices.
5) O(0,0), K(0, 12), L(4,0)
6) A(-7,0), O(0,0),B(0,-10)
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CF and EF are angle bisectors of ∆CDE.
Find each measure.
7)
The distance from F to CD
8)
m/ FED
C
17˚
F
54˚
42.1
D
G
E
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9) The designer of the Newtown High School pennant
wants the circle around the bear emblem to be as large as
possible. Draw a sketch to show where the center of the
circle should be located. Justify your sketch.
BEARS
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Let’s review
Bisector of Triangles
Since a triangle has three sides, it
has three perpendicular bisector.
When you construct the
perpendicular bisectors, you find
that they have an interesting
property.
CONFIDENTIAL
30
1
2
3
B
B
P
C
A
C
Draw a large
scalene acute
triangle ABC on a
piece of patty
paper.
A
Fold the
perpendicular
bisector of each
side.
CONFIDENTIAL
C
Label the point
where the three
perpendicular
bisectors
intersect as P.
31
When three or more lines intersect at one point,
the lines are said to be concurrent. The point of
concurrency is the point where they intersect.
In the construction, you saw that the three
perpendicular bisectors of a triangle are
concurrent. This point of concurrency is the
circumcenter of the triangle.
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32
Theorem 2.1
Circumcenter Theorem
B
The circumcenter of a triangle is
equidistant from the vertices of the
triangle.
P
PA = PB = PC
A
CONFIDENTIAL
C
33
The circumcenter can be inside the triangle, Out the triangle,
or on the triangle.
P
P
P
Acute triangle
Obtuse triangle
CONFIDENTIAL
Right triangle
34
The circumcenter of ∆ABC is the center of its
circumscribed circle. A circle that contains all the
vertices of a polygon is circumscribed about the
polygon.
B
P
A
C
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35
Circumcenter Theorem
A
l
n
Given: Line l , m, and n are the
perpendicular bisec tor of AB, BC,
and AC, respectively.
Prove: PA = PB = PC
B
Proof:
P is the circumcenter of ABC. Since P
lies on the perpendicular bisector of AB,
PA = PB by the Perpendicular Bisector
Theorem. Similarly, P also lies on t he
perpendicular bisector of BC, so PB = PC.
Therefore PA = PB = PC by the
Transit ive Propert y of Equality.
CONFIDENTIAL
P
C
m
36
Using Properties of
Perpendicular Bisector
H
KZ, LZ, and MZ are the perpendicular bisectors
of GHJ. Find HZ.
18.6
Z is the circumcenter of ∆GHJ. By the
Circumcenter Theorem, Z is equidistant from
the vertices of ∆GHJ.
K
HZ = GZ
Circumcenter Thm.
HZ = 19.9
Substitute 19.9 for GZ. 19.9
G
CONFIDENTIAL
Z
L
9.5
M
J
14.5
37
Finding the Circumcenter of a
Triangle.
Finding the circumcenter of ∆RSO with vertices
R(-6,0), S(0,4), and O (0,0).
Step 1 Graph the triangle.
y
X = -4
Y=2
S
(-3,2)
R 4
CONFIDENTIAL
O
x
Next page:
38
Step 2 Find equation for two perpendicular bisectors. Since
Two sides of the triangle lie along the axes, use the graph to
find the perpendicular bisectors of these two sides. The
perpendicular bisector of RO is x = -3, and the perpendicular
bisector of OS is y =2.
y
X = -4
Y=2
S
(-3,2)
R 4
CONFIDENTIAL
O
x
Next page:
39
Step 3 Find the intersection of the two equations.
The lines x = -3 and y = z intersect at (-3,2), the
circumcenter of ∆RSO.
y
X = -4
Y=2
S
(-3,2)
R 4
CONFIDENTIAL
O
x
40
Theorem 2.2
Incenter Theorem
The incenter of a triangle is equidistant from the
sides of the triangle.
B
PX = PY = PZ
Z
Y
P
A
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X
C
41
Unlike the circumcenter, the incenter is always inside the
triangle.
P
P
Acute triangle
Obtuse triangle
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P
Right triangle
42
The incenter is the center of the triangle’s inscribed
circle. A circle inscribed in a polygon intersects each
line that contains a side of the polygon at exactly one
point.
B
P
A
C
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Using Properties of Angle
Bisectors
K
A) The distance from V to KL
7.3
V is the incenter of ∆JKL. By the
Incenter Theorem, V is equidistant from
the sides of ∆JKL.
W
The distance from V to JK is 7.3.
V
106˚
So the distance from V to KL is also 7.3
J
19˚
L
Next page:
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44
B)
m/ Vkl
JV is the bisector of m/ kJL.
Substitute 19˚ for m / VJL. ∆
m/ kJL = 2(19˚) = 38˚
sum Thm.
m/ kJL + m/ JLK + m/ JKL = 180˚ Substitute the
7.3
given values.
38 + 106 + m/ JKL = 180˚
Subtract 144˚ from W
both sides.
m/ JKL = 36˚
K
m/ kJL = 2m / VJL
m/ Vkl = ½ m/ JKL
KV is the bisector of m/ JKL
m/ Vkl = ½ (36˚) = 18˚
V
106˚
Substitute 36˚ for
m/ JKL.
19˚
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L
45
Community Application
The city of Odessa will host a fireworks
display for the next Fourth of July
celebration. Draw a sketch to show where
the display should be positioned so that it is
the same distance from all three viewing
location A, B, and C on the map. Justify your
sketch.
A
Let the three viewing locations
be vertices of a triangle. By
the Circumcenter Theorem, the
circumcenter of the triangle is
equidistant from the vertices.
C
B
Next page:
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Trace the map. Draw the triangle formed by the
viewing locations. To find the circumcenter, find
the perpendicular bisectors of each side. The
position of the display is the circumcenter, F.
A
C
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F
B
47
You did a great job
today!
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