Chapter 5
Relationships Within 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
Midsegment
Base
Midsegment = ½ of Base
M = ½ b

23
46
M = ½ b
M = ½ • 46
M = 23

46
92
2•
M = ½ b
46= ½ b
92 = b

5x - 1
5•3-1 = 14
6x + 10
6•3+10 = 28
M = ½ b
5x - 1 = ½(6x + 10)
5x - 1 = 3x + 5
-3x -3x
2x - 1 = 5
+1 +1
2x = 6
2 2 x = 3

In ΔEFG, H, J, and K are midpoints. Find HJ, JK, and FG.
F
60
H J
HJ:
JK:
FG:
40
G
E K
100

AB = 10 and CD = 18. Find EB, BC, and AC
A
E B
C
D
EB:
BC:
AC:

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 its length
Example 1: Finding Lengths
In 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 +
NP +
+ = 60
= 60 x
NP =
M
24
P
22
Y Z
N

Find m<VUZ. Justify your answer.
X
65 °
U
Y
V
Z

In the diagram, ST and TU are midsegments of triangle PQR. Find PR and TU.

In the diagram, XZ and ZY are midsegments of triangle LMN. Find MN and ZY.

Dean 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 118 strides
128 strides
35 strides x
118 strides

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

D
B
C
If BD = DC, then we say that
D is equidistant from B and C.
Definition of an Equidistant Point
A point D is equidistant from B and C if and only if BD = DC.

R
U
M
T
V
S
Let TU be a perpendicular bisector of RS.
RT = TS
Then, what can you say about T, V and U?
RV = VS
RU = US
Theorem: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment.

Theorem 5-2:
Perpendicular Bisector
Thm.
Theorem 5-3: Converse of the Perpendicular Bisector
Thm.
If a point is on the perpendicular bisector of a segment, then it is equidistant form the endpoints of the segment.
If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

H
F G
Given: HF = HG
Conclusion: H lies on the perpendicular bisector of FG.

T
V
R S
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.
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.



A line, ray, or segment that is  to a side of the Δ at the midpoint of the side.
B l
A M
C

A Perpendicular bisector of a side does not have to start at a vertex. It will form a 90 ° angles and bisect the side.

Any point on the perpendicular bisector of a segment is equidistance from the endpoints of the segment.
C
A
AB is the perpendicular bisector of CD
D
B

This makes the Circumcenter an equidistance from the 3 vertices

Theorem 5-4:
Angle Bisector Thm.
Theorem 5-5:
Converse of the Angle
Bisector Thm.
If a point is on the bisector of an angle, then it is equidistant from the sides of the angle.
If a point in the interior of an angle is equidistant from the sides of the angle, then it is on the angle bisector.

Let AD be a bisector of  BAC,
P lie on AD,
PM  AB at M,
NP  AC at N.
A
M
B
N
C
Then P is equidistant from AB and AC.
P
D
Theorem: If a point lies on the bisector of an angle, then the point is equidistant from the sides of the angle.

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
C
A
D
3
B

CD is the perpendicular bisector of AB
Find CA and DB
C
5
A
6
D
B

Using the Angle Bisector Thm.
Find x, FB and FD in the diagram at the right.
Show steps to find x, FB and FD:
FD = Angle Bisector Thm.
7x – 35 = 2x + 5
A
2x + 5
B
F
7x - 35
C
D
E

a. According to the diagram, how far is K from ray EH? From ray ED?
D
C
K
10
H
2x
O
E
( X + 20) O

b. What can you conclude about ray EK?
D
C
K
10
H
2x
O
E
( X + 20) O

c. Find the value of x.
D
C
K
10
H
2x
O
E
( X + 20) O

d. Find m<DEH.
D
C
K
10
H
2x
O
E
( X + 20) O