Transcript 5.1 Perpendiculars and Bisectors Geometry Mrs. Spitz Fall 2004 Objectives: • Use properties of perpendicular bisectors • Use properties of angle bisectors to identify equal distances such.

5.1 Perpendiculars
and Bisectors
Geometry
Mrs. Spitz
Fall 2004
Objectives:
• Use properties of perpendicular bisectors
• Use properties of angle bisectors to
identify equal distances such as the
lengths of beams in a room truss.
Assignment:
• pp. 267-269 #1-26, 28, 32-35 all – Quiz
after 5.3
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
Perpendicular Bisector
Construction – pg. 264
1.
2.
3.
4.
5.
6.
If you cannot follow these directions, to go page 264 for
the instructions with pictures.
Draw a line. Any line about the middle of the page.
Place compass point at P. Draw an arc that intersects
line m twice. Label the intersections as A and B.
Use a compass setting greater than AP. Draw an arc
from A. With the same setting, draw an arc from B.
Label the intersection of the arcs as C.
Use a straightedge to draw CP. This line is
perpendicular to line m and passes through P.
Place this in your binder under “computer/lab work”
More about perpendicular
bisector construction
• You can measure CPA on your
construction to verify that the constructed
line is perpendicular to the given line m. In
the construction, CP  AB and PA = PB, so
CP is the perpendicular bisector of AB.
Equidistant
• A point is equidistant from two points if its
distance from each point is the same. In the
construction above, C is equidistant from A and B
because C was drawn so that CA = CB.
• Theorem 5.1 states that any point on the
perpendicular bisector CP in the construction is
equidistant from A and B, the endpoints of the
segment. The converse helps you prove that a
given point lies on a perpendicular bisector.
Theorem 5.1 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
Theorem 5.2: 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 ∆ABC
and ∆BPC are congruent using
the SAS Congruence
Postulate. Then show that CA
≅ CB.
• Exercise 28 asks you to write a
two-column proof of Theorem
5.1 using this plan. (This is part
of your homework)
C
A
P
perpendicular bisector
B
C
Given: CP is
perpendicular to
AB.
Prove: CA≅CB
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
A
Reasons:
1. Given
P
perpendicular bisector
B
C
Given: CP is
perpendicular to
AB.
Prove: CA≅CB
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
A
P
Reasons:
1. Given
2. Definition of
Perpendicular
bisector
perpendicular bisector
B
C
Given: CP is
perpendicular to
AB.
Prove: CA≅CB
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
A
P
Reasons:
1. Given
2. Definition of
Perpendicular
bisector
3. Given
perpendicular bisector
B
C
Given: CP is
perpendicular to
AB.
Prove: CA≅CB
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
A
Reasons:
P
B
perpendicular bisector
1. Given
2. Definition of
Perpendicular bisector
3. Given
4. Reflexive Prop.
Congruence.
C
Given: CP is
perpendicular to
AB.
Prove: CA≅CB
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
A
Reasons:
P
B
perpendicular bisector
1. Given
2. Definition of
Perpendicular bisector
3. Given
4. Reflexive Prop.
Congruence.
5. Definition right angle
C
Given: CP is
perpendicular to
AB.
Prove: CA≅CB
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
A
Reasons:
P
B
perpendicular bisector
1. Given
2. Definition of
Perpendicular bisector
3. Given
4. Reflexive Prop.
Congruence.
5. Definition right angle
6. SAS Congruence
C
Given: CP is
perpendicular to
AB.
Prove: CA≅CB
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
A
Reasons:
P
B
perpendicular bisector
1. Given
2. Definition of
Perpendicular bisector
3. Given
4. Reflexive Prop.
Congruence.
5. Definition right angle
6. SAS Congruence
7. CPCTC
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
Theorem 5.3 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
Theorem 5.3 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
4.3
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
32. Developing Proof
Given: D is in the
interior of ABC and is
equidistant from BA and
BC.
Prove: D lies on the
angle bisector of ABC.
A
B
D
C
A
Given: D is in the interior of ABC and is
equidistant from BA and BC.
Prove: D lies on the angle bisector of
ABC.
B
D
C
Statements:
1. D is in the interior of ABC.
2. D is ___?_ from BA and BC.
3. ____ = ____
4. DA  ____, ____  BC
5. __________
6. __________
7. BD ≅ BD
8. __________
9. ABD ≅ CBD
10.BD bisects ABC and point
D is on the bisector of ABC
Reasons:
1. Given
A
Given: D is in the interior of ABC and is
equidistant from BA and BC.
Prove: D lies on the angle bisector of
ABC.
B
D
C
Statements:
1. D is in the interior of ABC.
2. D is EQUIDISTANT from BA and BC.
3. ____ = ____
4. DA  ____, ____  BC
5. __________
6. __________
7. BD ≅ BD
8. __________
9. ABD ≅ CBD
10.BD bisects ABC and point D is
on the bisector of ABC
Reasons:
1. Given
2. Given
A
Given: D is in the interior of ABC and is
equidistant from BA and BC.
Prove: D lies on the angle bisector of
ABC.
B
D
C
Statements:
1. D is in the interior of ABC.
2. D is EQUIDISTANT from BA and BC.
3. DA = DC
4. DA  ____, ____  BC
5. __________
6. __________
7. BD ≅ BD
8. __________
9. ABD ≅ CBD
10.BD bisects ABC and point D is
on the bisector of ABC
Reasons:
1.
2.
3.
Given
Given
Def. Equidistant
A
Given: D is in the interior of ABC and is
equidistant from BA and BC.
Prove: D lies on the angle bisector of
ABC.
B
D
C
Statements:
1. D is in the interior of ABC.
2. D is EQUIDISTANT from BA and
BC.
3. DA = DC
4. DA  _BA_, __DC_  BC
5. __________
6. __________
7. BD ≅ BD
8. __________
9. ABD ≅ CBD
10.BD bisects ABC and point
D is on the bisector of ABC
Reasons:
1. Given
2. Given
3. Def. Equidistant
4. Def. Distance from point
to line.
A
Given: D is in the interior of ABC and is
equidistant from BA and BC.
Prove: D lies on the angle bisector of
ABC.
B
D
C
Statements:
1. D is in the interior of ABC.
2. D is EQUIDISTANT from BA and
BC.
3. DA = DC
4. DA  _BA_, __DC_  BC
5. DAB = 90°DCB = 90°
6. __________
7. BD ≅ BD
8. __________
9. ABD ≅ CBD
10.BD bisects ABC and point
D is on the bisector of ABC
Reasons:
1.
2.
3.
4.
5.
Given
Given
Def. Equidistant
Def. Distance from point to
line.
If 2 lines are , then they
form 4 rt. s.
A
Given: D is in the interior of ABC and is
equidistant from BA and BC.
Prove: D lies on the angle bisector of
ABC.
B
D
C
Statements:
1. D is in the interior of ABC.
2. D is EQUIDISTANT from BA and
BC.
3. DA = DC
4. DA  _BA_, __DC_  BC
5.
DAB and DCB are rt. s
6.
7.
8.
9.
10.
DAB = 90°DCB = 90°
BD ≅ BD
__________
ABD ≅ CBD
BD bisects ABC and point D is
on the bisector of ABC
Reasons:
1. Given
2. Given
3. Def. Equidistant
4. Def. Distance from point to
line.
5. If 2 lines are , then they
form 4 rt. s.
6. Def. of a Right Angle
A
Given: D is in the interior of ABC and is
equidistant from BA and BC.
Prove: D lies on the angle bisector of
ABC.
B
D
C
Statements:
1. D is in the interior of ABC.
2. D is EQUIDISTANT from BA and
BC.
3. DA = DC
4. DA  _BA_, __DC_  BC
5.
DAB and DCB are rt. s
6.
7.
8.
9.
10.
DAB = 90°DCB = 90°
BD ≅ BD
__________
ABD ≅ CBD
BD bisects ABC and point D is
on the bisector of ABC
Reasons:
1. Given
2. Given
3. Def. Equidistant
4. Def. Distance from point to
line.
5. If 2 lines are , then they
form 4 rt. s.
6. Def. of a Right Angle
7. Reflexive Property of Cong.
A
Given: D is in the interior of ABC and is
equidistant from BA and BC.
Prove: D lies on the angle bisector of
ABC.
B
D
C
Statements:
1. D is in the interior of ABC.
2. D is EQUIDISTANT from BA and
BC.
3. DA = DC
4. DA  _BA_, __DC_  BC
5.
DAB and DCB are rt. s
6.
7.
8.
9.
10.
DAB = 90°DCB = 90°
BD ≅ BD
∆ABD ≅ ∆CBD
ABD ≅ CBD
BD bisects ABC and point D is
on the bisector of ABC
Reasons:
1. Given
2. Given
3. Def. Equidistant
4. Def. Distance from point to
line.
5. If 2 lines are , then they
form 4 rt. s.
6. Def. of a Right Angle
7. Reflexive Property of Cong.
8. HL Congruence Thm.
A
Given: D is in the interior of ABC and is
equidistant from BA and BC.
Prove: D lies on the angle bisector of
ABC.
B
D
C
Statements:
1. D is in the interior of ABC.
2. D is EQUIDISTANT from BA and
BC.
3. DA = DC
4. DA  _BA_, __DC_  BC
5.
DAB and DCB are rt. s
6.
7.
8.
9.
10.
DAB = 90°DCB = 90°
BD ≅ BD
∆ABD ≅ ∆CBD
ABD ≅ CBD
BD bisects ABC and point D is
on the bisector of ABC
Reasons:
1. Given
2. Given
3. Def. Equidistant
4. Def. Distance from point to
line.
5. If 2 lines are , then they
form 4 rt. s.
6. Def. of a Right Angle
7. Reflexive Property of Cong.
8. HL Congruence Thm.
9. CPCTC
A
Given: D is in the interior of ABC and is
equidistant from BA and BC.
Prove: D lies on the angle bisector of
ABC.
B
D
C
Statements:
1. D is in the interior of ABC.
2. D is EQUIDISTANT from BA and
BC.
3. DA = DC
4. DA  _BA_, __DC_  BC
5.
DAB and DCB are rt. s
6.
7.
8.
9.
10.
DAB = 90°DCB = 90°
BD ≅ BD
∆ABD ≅ ∆CBD
ABD ≅ CBD
BD bisects ABC and point D is
on the bisector of ABC
Reasons:
1. Given
2. Given
3. Def. Equidistant
4. Def. Distance from point to
line.
5. If 2 lines are , then they
form 4 rt. s.
6. Def. of a Right Angle
7. Reflexive Property of Cong.
8. HL Congruence Thm.
9. CPCTC
10. Angle Bisector Thm.