Transcript 5.1 & 5.2

5.1 Midsegments of
Triangles
Chapter 5
Relationships 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
Midsegment
Base
Midsegment = ½ of Base
M=½b
Find the length of the midsegment.
23
46
M=½b
M = ½ • 46
M = 23
Find the length of the base.
46
92
M = ½b
2•46=2• ½ b
92 = b
Find the length of the midsegment and
the base.
5•3-1 = 14
5x - 1
6x + 10
6•3+10 = 28
M = ½b
5x - 1 = ½(6x + 10)
5x - 1 = 3x + 5
-3x
-3x
2x - 1 = 5
+1 +1
2x = 6
2
2
x=3
5.1 Midsegments of Triangles
In ΔEFG, H, J, and K are midpoints. Find HJ, JK, and FG.
F
HJ:
JK:
60
H
FG:
J
40
E
K
100
G
5.1 Midsegments of Triangles
AB = 10 and CD = 18. Find EB, BC, and AC
A
E
D
EB:
BC:
AC:
B
C
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
60. Find NP and YZ.
MNP is
Because the perimeter is 60, you can find NP.
NP + MN + MP = 60 (Definition of Perimeter)
NP +
NP +
+
= 60
x
= 60
NP =
24
M
P
22
Y
N
Z
5.1 Midsegments
Find m<VUZ. Justify your answer.
X
65°
U
Z
Y
V
Example 1
In the diagram, ST and TU are midsegments of
triangle PQR. Find PR and TU.
16 ft
PR = ________
5 ft
TU = ________
Example 2
In the diagram, XZ and ZY are midsegments
of triangle LMN. Find MN and ZY.
53 cm
MN = ________
14 cm
ZY = ________
5.1 Midsegments
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
118 strides
35 strides
x
(5.2) Bisectors in Triangles
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.
T
V
M
R
S
U
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.
Theorems
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.
 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
l
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.
Circumcenter
Any point on the perpendicular bisector
of a segment is equidistance from the
endpoints of the segment.
A
C
B
AB is the
perpendicular
bisector of CD
D
This makes the Circumcenter an
equidistance from the 3 vertices
Theorems
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 If a point in the interior of
it is equidistant from the an angle is equidistant from
sides of the angle.
the sides of the angle, then
it is on the angle bisector.
Let AD be a bisector of BAC,
M
B
P lie on AD,
PM  AB at M,
P
A
D
NP  AC at N.
N
C
Then P is equidistant from AB and AC.
Theorem: If a point lies on the bisector of an
angle, then the point is equidistant from the
sides of the angle.
Key Concepts
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
D
3
A
B
CD is the perpendicular bisector of AB
Find CA and DB
C
5
A
B
6
D
Example
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 =
7x – 35 = 2x + 5
A
2x + 5
B
F
C
Angle Bisector Thm.
D
7x - 35
E
Quick Check
a. According to the diagram, how far is K
from ray EH? From ray ED?
2xO
D
E
C
(X + 20)O
K
10
H
Quick Check
b. What can you conclude about ray EK?
2xO
D
E
C
(X + 20)O
K
10
H
Quick Check
c. Find the value of x.
2xO
D
E
C
(X + 20)O
K
10
H
Quick Check
d. Find m<DEH.
2xO
D
E
C
(X + 20)O
K
10
H