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Geometry Bisector of Triangles. 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) 3) Z(-8,-2). yes yes no. Bisector of Triangles. - PowerPoint PPT Presentation
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CONFIDENTIAL 1
Geometry
Bisector of Triangles
CONFIDENTIAL 2
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) 3) Z(-8,-2)
1) yes2) yes3) no
CONFIDENTIAL 3
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.
Bisector of Triangles
CONFIDENTIAL 4
1 2 3
A
B
C
A
C
A
B
C
P
Draw a large scalene acute
triangle ABC on a piece of patty
paper.
Fold the perpendicular
bisector of each side.
Label the point where the three perpendicular
bisectors intersect as P.
CONFIDENTIAL 5
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.
CONFIDENTIAL 6
Theorem 2.1
Circumcenter Theorem
The circumcenter of a triangle is equidistant from the vertices of the triangle.
PA = PB = PC
P
A
B
C
CONFIDENTIAL 7
The circumcenter can be inside the triangle, Out the triangle, or on the triangle.
P
P
P
Acute triangle Obtuse triangle Right triangle
CONFIDENTIAL 8
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.
A
B
C
P
CONFIDENTIAL 9
Circumcenter Theorem
Given: Line l , m, and n are the
perpendicular bisector of AB, BC, and AC, respectively. Prove: PA = PB = PCProof: 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 the perpendicular bisector of BC, so PB = PC. Therefore PA = PB = PC by the Transitive Property of Equality.
n
P
A
m
CB
l
CONFIDENTIAL 10
Using Properties of Perpendicular Bisector
Z is the circumcenter of ∆GHJ. By the Circumcenter Theorem, Z is equidistant from the vertices of ∆GHJ.
HZ = GZ Circumcenter Thm.
HZ = 19.9 Substitute 19.9 for GZ.
KZ, LZ, and MZ are the perpendicular bisectors of GHJ. Find HZ.
H
JG
K L
19.9
Z
M
18.6
14.5
9.5
CONFIDENTIAL 11
Now you try!
1) Use the diagram. Find each length.
a) GM b) GK c) JZ
H
JG
K L
19.9
Z
M
18.6
14.5
9.5
1a) 14.51b) 18.61c) 19.9
CONFIDENTIAL 12
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.
O
S
X = -4
y
x4R
Y = 2 (-3,2)
Next page:
CONFIDENTIAL 13
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.
O
S
X = -4
y
x4R
Y = 2 (-3,2)
Next page:
CONFIDENTIAL 14
O
S
X = -4
y
x4R
Y = 2 (-3,2)
Step 3 Find the intersection of the two equations.
The lines x = -3 and y = z intersect at (-3,2), the circumcenter of ∆RSO.
CONFIDENTIAL 15
Now you try!
2) Find the circumcenter of ∆GOH with vertices
G(0,-9), O(0,0), and H(8,0).
2) 29
CONFIDENTIAL 16
Theorem 2.2 Incenter Theorem
The incenter of a triangle is equidistant from the sides of the triangle.
PX = PY = PZ
P
ZY
XA
B
C
CONFIDENTIAL 17
Unlike the circumcenter, the incenter is always inside the triangle.
P PP
Acute triangle Obtuse triangleRight triangle
CONFIDENTIAL 18
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.
P
A
B
C
CONFIDENTIAL 19
Using Properties of Angle Bisectors
7.3
106˚
19˚
J
W
L
V
K
A) The distance from V to KL
V is the incenter of ∆JKL. By the Incenter Theorem, V is equidistant from the sides of ∆JKL.
The distance from V to JK is 7.3.
So the distance from V to KL is also 7.3
Next page:
CONFIDENTIAL 20
7.3
106˚
19˚
W
L
V
K
B) m/ Vkl
m/ kJL = 2m / VJL
m/ kJL = 2(19˚) = 38˚
m/ kJL + m/ JLK + m/ JKL = 180˚
38 + 106 + m/ JKL = 180˚
m/ JKL = 36˚
m/ Vkl = ½ m/ JKL
m/ Vkl = ½ (36˚) = 18˚
JV is the bisector of m/ kJL. Substitute 19˚ for m / VJL. ∆ sum Thm.
Substitute 36˚ for m/ JKL.
Substitute the given values. Subtract 144˚ from both sides.
KV is the bisector of m/ JKL
CONFIDENTIAL 21
Now you try!
3) QX and RX are angle bisectors ∆PQR. Find each measure.
a) The distance from X to PQ
b) m/ PQX
P
Q
R
X
Y 19.212˚
3a) 14.53b) 18.6
CONFIDENTIAL 22
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.
Let the three viewing locations be vertices of a triangle. By the Circumcenter Theorem, the circumcenter of the triangle is equidistant from the vertices.
Next page:
A
BC
CONFIDENTIAL 23
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
BC
F
CONFIDENTIAL 24
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.
Centerville Anverue
Kin
g B
ou
levard
Third Street
CONFIDENTIAL 25
Now some practice problems for you!
CONFIDENTIAL 26
SN, TN, and VN are the perpendicular bisectors of ∆PQR. Find each length.
1) NR 2) RV
3) TR 4) QN N
Q
R
T
P
S
5.47 V
4.03
3.95
5.64
Assessment
CONFIDENTIAL 27
5) O(0,0), K(0, 12), L(4,0)
6) A(-7,0), O(0,0),B(0,-10)
Find the circumcenter of a triangle with the given vertices.
CONFIDENTIAL 28
CF and EF are angle bisectors of ∆CDE.
Find each measure.
7) The distance from F to CD
8) m/ FED
C D
E
G
F42.1
54˚
17˚
CONFIDENTIAL 29
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
CONFIDENTIAL 30
Let’s review
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.
Bisector of Triangles
CONFIDENTIAL 31
1 2 3
A
B
C
A
C
A
B
C
P
Draw a large scalene acute
triangle ABC on a piece of patty
paper.
Fold the perpendicular
bisector of each side.
Label the point where the three perpendicular
bisectors intersect as P.
CONFIDENTIAL 32
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.
CONFIDENTIAL 33
Theorem 2.1 Circumcenter Theorem
The circumcenter of a triangle is equidistant from the vertices of the triangle.
PA = PB = PC
P
A
B
C
CONFIDENTIAL 34
The circumcenter can be inside the triangle, Out the triangle, or on the triangle.
P
P
P
Acute triangle Obtuse triangle Right triangle
CONFIDENTIAL 35
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.
A
B
C
P
CONFIDENTIAL 36
Circumcenter Theorem
Given: Line l , m, and n are the
perpendicular bisector of AB, BC, and AC, respectively. Prove: PA = PB = PCProof: 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 the perpendicular bisector of BC, so PB = PC. Therefore PA = PB = PC by the Transitive Property of Equality.
n
P
A
m
CB
l
CONFIDENTIAL 37
Using Properties of Perpendicular Bisector
Z is the circumcenter of ∆GHJ. By the Circumcenter Theorem, Z is equidistant from the vertices of ∆GHJ.
HZ = GZ Circumcenter Thm.
HZ = 19.9 Substitute 19.9 for GZ.
KZ, LZ, and MZ are the perpendicular bisectors of GHJ. Find HZ.
H
JG
K L
19.9
Z
M
18.6
14.5
9.5
CONFIDENTIAL 38
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.
O
S
X = -4
y
x4R
Y = 2 (-3,2)
Next page:
CONFIDENTIAL 39
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.
O
S
X = -4
y
x4R
Y = 2 (-3,2)
Next page:
CONFIDENTIAL 40
O
S
X = -4
y
x4R
Y = 2 (-3,2)
Step 3 Find the intersection of the two equations.
The lines x = -3 and y = z intersect at (-3,2), the circumcenter of ∆RSO.
CONFIDENTIAL 41
Theorem 2.2 Incenter Theorem
The incenter of a triangle is equidistant from the sides of the triangle.
PX = PY = PZ
P
ZY
XA
B
C
CONFIDENTIAL 42
Unlike the circumcenter, the incenter is always inside the triangle.
P PP
Acute triangle Obtuse triangleRight triangle
CONFIDENTIAL 43
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.
P
A
B
C
CONFIDENTIAL 44
Using Properties of Angle Bisectors
7.3
106˚
19˚
J
W
L
V
K
A) The distance from V to KL
V is the incenter of ∆JKL. By the Incenter Theorem, V is equidistant from the sides of ∆JKL.
The distance from V to JK is 7.3.
So the distance from V to KL is also 7.3
Next page:
CONFIDENTIAL 45
7.3
106˚
19˚
W
L
V
K
B) m/ Vkl
m/ kJL = 2m / VJL
m/ kJL = 2(19˚) = 38˚
m/ kJL + m/ JLK + m/ JKL = 180˚
38 + 106 + m/ JKL = 180˚
m/ JKL = 36˚
m/ Vkl = ½ m/ JKL
m/ Vkl = ½ (36˚) = 18˚
JV is the bisector of m/ kJL. Substitute 19˚ for m / VJL. ∆ sum Thm.
Substitute 36˚ for m/ JKL.
Substitute the given values. Subtract 144˚ from both sides.
KV is the bisector of m/ JKL
CONFIDENTIAL 46
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.
Let the three viewing locations be vertices of a triangle. By the Circumcenter Theorem, the circumcenter of the triangle is equidistant from the vertices.
Next page:
A
BC
CONFIDENTIAL 47
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
BC
F
CONFIDENTIAL 48
You did a great job today!
