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!