Transcript 5.1 Perpendiculars and Bisectors

5.1 Perpendiculars
and Bisectors
Geometry
Mrs. Spitz
Fall 2004
Use Properties of
Perpendicular Bisectors
• In lesson 1.5, you learned that
a segment bisector intersects a
segment at its midpoint. A
segment, ray, line, or plane that
is perpendicular to a segment
at its midpoint is called a
perpendicular bisector.
The
construction on pg. 264 shows
how to draw a line that is
perpendicular to a given line or
segment at a point P. You can
use this method to construct a
perpendicular bisector or a
segment as described in the
activity.
C
Given segment
A
CP is a  bisector of AB
P
perpendicular bisector
B
Equidistant
• A point is equidistant from two points if its
distance from each point is the same.
Perpendicular Bisector
Theorem
If a point is on the
perpendicular bisector
of a segment, then it is
equidistant from the
endpoints of the
segment.
C
A
P
perpendicular bisector
If
CP
is
the
perpendicular bisector
of AB, then CA = CB.
B
Converse of the Perpendicular
Bisector Theorem
If a point is equidistant
from the endpoints of a
segment, then it is on
the perpendicular
bisector of the segment.
If DA = DB, then D lies
on the perpendicular
bisector of AB.
C
P
A
D is on CP
B
D
Plan for Proof of Theorem 5.1
• Refer to the diagram for Theorem
5.1. Suppose that you are given
that CP is the perpendicular
bisector of AB. Show that right
triangles ∆APC and ∆BPC are
congruent
using
the
SAS
Congruence Postulate.
Then
show that CA ≅ CB.
C
A
P
perpendicular bisector
B
Given: CP is
perpendicular to
AB. AP≅BP
C
Prove: CA≅CB
A
P
perpendicular bisector
Statements:
1. CP is perpendicular
bisector of AB.
2. CP  AB
3. AP ≅ BP
4. CP ≅ CP
5. CPB ≅ CPA
6. ∆APC ≅ ∆BPC
7. CA ≅ CB
Reasons:
1. Given
B
Given: CP is
perpendicular to
AB. AP≅BP
C
Prove: CA≅CB
A
P
perpendicular bisector
Statements:
1. CP is perpendicular
bisector of AB.
2. CP  AB
3. AP ≅ BP
4. CP ≅ CP
5. CPB ≅ CPA
6. ∆APC ≅ ∆BPC
7. CA ≅ CB
Reasons:
1. Given
2. Definition of
Perpendicular
bisector
B
Given: CP is
perpendicular to
AB. AP≅BP
C
Prove: CA≅CB
A
P
perpendicular bisector
Statements:
1. CP is perpendicular
bisector of AB.
2. CP  AB
3. AP ≅ BP
4. CP ≅ CP
5. CPB ≅ CPA
6. ∆APC ≅ ∆BPC
7. CA ≅ CB
Reasons:
1. Given
2. Definition of
Perpendicular
bisector
3. Given
B
Given: CP is
perpendicular to
AB. AP≅BP
C
Prove: CA≅CB
A
P
B
perpendicular bisector
Statements:
1. CP is perpendicular
bisector of AB.
2. CP  AB
3. AP ≅ BP
4. CP ≅ CP
5. CPB ≅ CPA
6. ∆APC ≅ ∆BPC
7. CA ≅ CB
Reasons:
1. Given
2. Definition of
Perpendicular bisector
3. Given
4. Reflexive Prop.
Congruence.
Given: CP is
perpendicular to
AB. AP≅BP
C
Prove: CA≅CB
A
P
B
perpendicular bisector
Statements:
1. CP is perpendicular
bisector of AB.
2. CP  AB
3. AP ≅ BP
4. CP ≅ CP
5. CPB ≅ CPA
6. ∆APC ≅ ∆BPC
7. CA ≅ CB
Reasons:
1. Given
2. Definition of
Perpendicular bisector
3. Given
4. Reflexive Prop.
Congruence.
5. Definition right angle
Given: CP is
perpendicular to
AB. AP≅BP
C
Prove: CA≅CB
A
P
B
perpendicular bisector
Statements:
1. CP is perpendicular
bisector of AB.
2. CP  AB
3. AP ≅ BP
4. CP ≅ CP
5. CPB ≅ CPA
6. ∆APC ≅ ∆BPC
7. CA ≅ CB
Reasons:
1. Given
2. Definition of
Perpendicular bisector
3. Given
4. Reflexive Prop.
Congruence.
5. Definition right angle
6. SAS Congruence
Given: CP is
perpendicular to
AB. AP≅BP
C
Prove: CA≅CB
A
P
B
perpendicular bisector
Statements:
1. CP is perpendicular
bisector of AB.
2. CP  AB
3. AP ≅ BP
4. CP ≅ CP
5. CPB ≅ CPA
6. ∆APC ≅ ∆BPC
7. CA ≅ CB
Reasons:
1. Given
2. Definition of
Perpendicular bisector
3. Given
4. Reflexive Prop.
Congruence.
5. Definition right angle
6. SAS Congruence
7. CPOCTAC
Ex. 1 Using
Perpendicular Bisectors
•In the diagram MN is
the perpendicular
bisector of ST.
a. What segment
lengths in the
diagram are equal?
b. Explain why Q is on
MN.
T
12
M
N
Q
12
S
Ex. 1 Using
Perpendicular Bisectors
a. What segment
lengths in the
diagram are equal?
Solution: MN bisects ST, so
NS = NT. Because M is
on the perpendicular
M
bisector of ST, MS = MT.
(By Theorem 5.1). The
diagram shows that QS =
QT = 12.
T
12
N
Q
12
S
Ex. 1 Using
Perpendicular Bisectors
b.Explain why Q is on
MN.
T
12
Solution: QS = QT, so
Q is equidistant from S
and T. By Theorem 5.2, M
Q is on the
perpendicular bisector
of ST, which is MN.
N
Q
12
S
Using Properties of
Angle Bisectors
• The distance from a point to
a line is defined as the
length of the perpendicular
segment from the point to
the line. For instance, in the
diagram shown, the distance
between the point Q and the
line m is QP.
Q
P
Using Properties of
Angle Bisectors
• When a point is the same
distance from one line as it is
from another line, then the
point is equidistant from the
two lines (or rays or
segments). The theorems in
the next few slides show that
a point in the interior of an
angle is equidistant from the
sides of the angle if and only
if the point is on the bisector
of an angle.
Q
P
Angle Bisector Theorem
If a point is on the
bisector of an angle,
then it is equidistant
from the two sides of
the angle.
If mBAD = mCAD,
then DB = DC
B
A
D
C
Converse of the Angle
Bisector Theorem
If a point is in the interior
of an angle and is
equidistant from the
sides of the angle, then
it lies on the bisector of
the angle.
If DB = DC, then mBAD
= mCAD.
B
A
D
C
Ex. 2: Proof of Theorem
5.3
Given: D is on the
bisector of BAC. DB
AB, DC  AC.
Prove: DB = DC
Plan for Proof: Prove
that ∆ADB ≅ ∆ADC.
Then conclude that DB
≅DC, so DB = DC.
B
D
A
C
Paragraph Proof
By definition of an angle
bisector, BAD ≅ CAD.
Because ABD and ACD
are right angles, ABD ≅
ACD. By the Reflexive
Property of Congruence, AD
≅ AD. Then ∆ADB ≅ ∆ADC
by the AAS Congruence
Theorem. By CPCTC, DB ≅
DC. By the definition of
congruent segments DB =
DC.
B
D
A
C
Ex. 3: Using Angle
Bisectors
Roof Trusses: Some roofs
are built with wooden
trusses that are assembled
in a factory and shipped to
the building site. In the
diagram of the roof trusses
shown, you are given that
AB bisects CAD and that
ACB and ADB are right
angles. What can you say
about BC and BD?
C A D
O
M
L
G
B
H
K
N
P
SOLUTION:
Because BC and BD meet
AC and AD at right angles,
they
are
perpendicular
segments to the sides of
CAD. This implies that
their
lengths
represent
distances from the point B
to AC and AD. Because
point B is on the bisector of
CAD, it is equidistant from
the sides of the angle.
So, BC = BD, and you can
conclude that BC ≅ BD.
C A D
O
M
L
G
B
H
K
N
P