5.1 Midsegments of Triangles

Chapter 5Relationships Within Triangles

5.1 Midsegments of Triangles Theorem 5-1 Triangle Midsegment Theorem

If a segment joins the midpoints of two sides of a triangle, then the segment is parallel to the

third side, and is half of its length

Midsegment

x

2x

Midsegment of a Triangle segment whose endpoints are the midpoints of two sides of the triangle

Base

Midsegment

Midsegment = ½ of Base

M = ½ b

46

M = ½ bM = ½ • 46

M = 23

23

Find the length of the midsegment.

Find the length of the base.

46 M = ½ b46= ½ b2• 2•92 = b92

Find the length of the midsegment and the base.

5x - 1

6x + 10

M = ½ b

5x - 1 = ½(6x + 10)

5x - 1 = 3x + 5-3x -3x2x - 1 = 5

+1 +12x = 6

2 2x = 3

5•3-1 = 14

6•3+10= 28

5.1 Midsegments of TrianglesIn ΔEFG, H, J, and K are midpoints. Find HJ, JK, and FG.

F

K G

JH

E

60

100

40

HJ:

JK:

FG:

5.1 Midsegments of TrianglesAB = 10 and CD = 18. Find EB, BC, and AC

A

BE

D C

EB:

BC:

AC:

Theorem 5-1: Triangle Midsegment TheoremIf a segment joins the midpoints of two sides of a triangle, then the segment is parallel to the third side, and is half its length

Example 1: Finding LengthsIn XYZ, M, N and P are the midpoints. The Perimeter of MNP is

60. Find NP and YZ.

Because the perimeter is 60, you can find NP.

NP + MN + MP = 60 (Definition of Perimeter)

NP + + = 60

NP + = 60

NP = 24

22

x

P

Y

M

N Z

5.1 MidsegmentsFind m<VUZ. Justify your answer.

65°

X

Y V

Z

U

Example 1In the diagram, ST and TU are midsegments of

triangle PQR. Find PR and TU.

PR = ________ TU = ________16 ft 5 ft

Example 2In the diagram, XZ and ZY are midsegments

of triangle LMN. Find MN and ZY.

MN = ________ ZY = ________53 cm 14 cm

5.1 MidsegmentsDean plans to swim the length of the lake, as shown

in the photo. He counts the distances shown by counting 3ft strides. How far would Dean swim?

35 strides

35 strides

118 strides

118 strides128 strides

x

(5.2) Bisectors in Triangles

What will we be learning today? Use properties of perpendicular bisectors and angle bisectors.

Definition of an Equidistant PointA point D is equidistant from B and C if and only if BD = DC.

D

BC

If BD = DC, then we say that

D is equidistant from B and C.

Theorem: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment.

R S

T

V

U

Let TU be a perpendicular bisector of RS.

Then, what can you say about T, V and U?

RT = TS

RV = VS

RU = US

M

Theorem 5-2: Perpendicular Bisector Thm.

If a point is on the perpendicular bisector of a segment, then it is equidistant form the endpoints of the segment.

Theorem 5-3: Converse of the Perpendicular Bisector Thm.

If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

Theorems

F G

H

Given: HF = HG

Conclusion: H lies on the perpendicular bisector of FG.

Theorem: If two points and a segment lie on the same plane and each of the two points are equidistant from the endpoints of the segment, then the line joining the points is

the perpendicular bisector of the segment.

R S

T

V

If T is equidistant from R and S and similarly, V is equidistant from R and S, then what can we say about TV?

TV is the perpendicular bisector of RS.

bisector of a Δ A line, ray, or segment that is to a side

of the Δ at the midpoint of the side.

B

A M C

l

A Perpendicular bisector of a side does not A Perpendicular bisector of a side does not have to start at a vertex. It will formhave to start at a vertex. It will form

a a 90° angles90° angles and bisectand bisect the side. the side.

CircumcenterCircumcenter

Any point on the Any point on the perpendicular bisectorperpendicular bisectorof a segment is equidistance from theof a segment is equidistance from the

endpoints of the segment.endpoints of the segment.

AA

BB

CC DD

AB is the AB is the perpendicularperpendicularbisector of CDbisector of CD

This makes the This makes the CircumcenterCircumcenter an anequidistance from the 3 verticesequidistance from the 3 vertices

Theorem 5-4: Angle Bisector Thm.

If a point is on the bisector of an angle, then it is equidistant from the sides of the angle.

Theorem 5-5: Converse of the Angle Bisector Thm.

If a point in the interior of an angle is equidistant from the sides of the angle, then it is on the angle bisector.

Theorems

D

Theorem: If a point lies on the bisector of an angle, then the point is equidistant from the sides of the angle.

Let AD be a bisector of BAC,

P lie on AD,

PM AB at M,

NP AC at N.

A

B

C

M

N

P

Then P is equidistant from AB and AC.

The distance from a point to a line is the length of the perpendicular segment from the point to the line.

Example: D is 3 in. from line AB and line AC

Key Concepts

C

B

D

A3

CC

AA BB

DD

55

66

is the perpendicular bisector of

Find and

CD AB

CA DB

Using the Angle Bisector Thm. Find x, FB and FD in the diagram at the right.

Example

7x - 35

A

E

F

CD

B 2x + 5

Show steps to find x, FB and FD:

FD = Angle Bisector Thm.

7x – 35 = 2x + 5

Quick Check

a. According to the diagram, how far is K from ray EH? From ray ED?

H

D

CE

(X + 20)O

2xO

K

10

Quick Check

b. What can you conclude about ray EK?

H

D

CE

(X + 20)O

2xO

K

10

Quick Check

c. Find the value of x.

H

D

CE

(X + 20)O

2xO

K

10

Quick Check

d. Find m<DEH.

H

D

CE

(X + 20)O

2xO

K

10