5.1 midsegments of triangles
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5.1 midsegments of triangles
GeometryMrs. SpitzFall 2004
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Objectives:
• Use properties of midsegments to solve problems
• Use properties of perpendicular bisectors
• Use properties of angle bisectors to identify equal distances such as the lengths of beams in a room truss.
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Midsegments of triangles
• Turn in your book to page 243. In the blue box is an investigation.
• Complete the investigation, we’ll share our conjectures in 3 minutes.
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Midsegment
• A segment that connects the midpoints of two sides
• 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 it’s length.
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Example 1
• In triangle XYZ, M, N, and P are midpoints. The perimeter of triangle MNP is 60. Find NP and YZ.
Y
X
Z
MP
N
22
24
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Example 2
• In triangle DEF, A, B, and C are midpoints. Name pairs of parallel segments.
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• Check for understanding. – AB= 10 and CD = 18. Find EB, BC, and AC.
• Critical Thinking– Find m<VUZ. Justify your answer
D
A
E B
C
=
= ||
V
X
Y
U
Z=
=
||
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Real world connection
• CD is a new bridge being built over a lake as shown. Find the length of the bridge.
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Assignment:
• Page 246 #1-20 we’ll go over them then you’ll turn in tomorrow #’s 21-36; 47-55
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5.2 Bisectors in Triangles
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Objectives:
• Use properties of midsegments to solve problems
• Use properties of perpendicular bisectors
• Use properties of angle bisectors to identify equal distances such as the lengths of beams in a room truss.
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Given segment
perpendicular bisector
PA B
C
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.
CP is a bisector of AB
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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.
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Perpendicular bisector theorem
• If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
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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.
If CP is the perpendicular bisector of AB, then CA = CB.
Given segment
perpendicular bisector
PA B
C
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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.
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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.
D is on CP
P
A B
C
D
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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.
Given segment
perpendicular bisector
PA B
C
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Statements:1. CP is perpendicular
bisector of AB.2. CP AB3. AP ≅ BP4. CP ≅ CP5. CPB ≅ CPA6. ∆APC ≅ ∆BPC7. CA ≅ CB
Reasons:1. Given
Given: CP is perpendicular to AB.
Prove: CA≅CBperpendicular bisector
PA B
C
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Statements:1. CP is perpendicular
bisector of AB.2. CP AB3. AP ≅ BP4. CP ≅ CP5. CPB ≅ CPA6. ∆APC ≅ ∆BPC7. CA ≅ CB
Reasons:1. Given2. Definition of
Perpendicular bisector
Given: CP is perpendicular to AB.
Prove: CA≅CBperpendicular bisector
PA B
C
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Statements:1. CP is perpendicular
bisector of AB.2. CP AB3. AP ≅ BP4. CP ≅ CP5. CPB ≅ CPA6. ∆APC ≅ ∆BPC7. CA ≅ CB
Reasons:1. Given2. Definition of
Perpendicular bisector
3. Given
Given: CP is perpendicular to AB.
Prove: CA≅CBperpendicular bisector
PA B
C
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Statements:1. CP is perpendicular
bisector of AB.2. CP AB3. AP ≅ BP4. CP ≅ CP5. CPB ≅ CPA6. ∆APC ≅ ∆BPC7. CA ≅ CB
Reasons:1. Given2. Definition of
Perpendicular bisector3. Given4. Reflexive Prop.
Congruence.
Given: CP is perpendicular to AB.
Prove: CA≅CBperpendicular bisector
PA B
C
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Statements:1. CP is perpendicular
bisector of AB.2. CP AB3. AP ≅ BP4. CP ≅ CP5. CPB ≅ CPA6. ∆APC ≅ ∆BPC7. CA ≅ CB
Reasons:1. Given2. Definition of
Perpendicular bisector3. Given4. Reflexive Prop.
Congruence.5. Definition right angle
Given: CP is perpendicular to AB.
Prove: CA≅CBperpendicular bisector
PA B
C
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Statements:1. CP is perpendicular
bisector of AB.2. CP AB3. AP ≅ BP4. CP ≅ CP5. CPB ≅ CPA6. ∆APC ≅ ∆BPC7. CA ≅ CB
Reasons:1. Given2. Definition of
Perpendicular bisector3. Given4. Reflexive Prop.
Congruence.5. Definition right angle6. SAS Congruence
Given: CP is perpendicular to AB.
Prove: CA≅CBperpendicular bisector
PA B
C
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Statements:1. CP is perpendicular
bisector of AB.2. CP AB3. AP ≅ BP4. CP ≅ CP5. CPB ≅ CPA6. ∆APC ≅ ∆BPC7. CA ≅ CB
Reasons:1. Given2. Definition of
Perpendicular bisector3. Given4. Reflexive Prop.
Congruence.5. Definition right angle6. SAS Congruence7. CPCTC
Given: CP is perpendicular to AB.
Prove: CA≅CBperpendicular bisector
PA B
C
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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.
12
12
N
T
S
QM
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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 bisector of ST, MS = MT. (By Theorem 5.1). The diagram shows that QS = QT = 12.
12
12
N
T
S
QM
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Ex. 1 Using Perpendicular Bisectors
b.Explain why Q is on MN.
Solution: QS = QT, so Q is equidistant from S and T. By Theorem 5.2, Q is on the perpendicular bisector of ST, which is MN.
12
12
N
T
S
QM
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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 mBAD = mCAD, then DB = DC
B
A
C
D
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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 mBAD = mCAD.
B
A
C
D
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Ex. 2: Proof of angle bisector theorem
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
A
C
D
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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
A
C
D
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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
C
D
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Statements:1.D is in the interior of ABC.2.D is ___?_ from BA and BC.3.____ = ____4.DA ____, ____ BC
5.__________6.__________7.BD ≅ BD8.__________9. ABD ≅ CBD10.BD bisects ABC and
point D is on the bisector of ABC
A
B
C
D
Reasons:
1. Given
Given: D is in the interior of ABC and is equidistant from BA and BC.Prove: D lies on the angle bisector of ABC.
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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 ≅ BD8. __________9. ABD ≅ CBD10.BD bisects ABC and point D
is on the bisector of ABC
A
B
C
D
Reasons:
1. Given
2. Given
Given: D is in the interior of ABC and is equidistant from BA and BC.Prove: D lies on the angle bisector of ABC.
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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 ≅ BD8. __________9. ABD ≅ CBD10.BD bisects ABC and point D
is on the bisector of ABC
A
B
C
D
Reasons:1. Given
2. Given
3. Def. Equidistant
Given: D is in the interior of ABC and is equidistant from BA and BC.Prove: D lies on the angle bisector of ABC.
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Statements:1.D is in the interior of ABC.2.D is EQUIDISTANT from BA and
BC.3.DA = DC4.DA _BA_, __DC_ BC5.__________6.__________7.BD ≅ BD8.__________9. ABD ≅ CBD10.BD bisects ABC and
point D is on the bisector of ABC
A
B
C
D
Reasons:
1. Given
2. Given
3. Def. Equidistant
4. Def. Distance from point to line.
Given: D is in the interior of ABC and is equidistant from BA and BC.Prove: D lies on the angle bisector of ABC.
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Statements:1.D is in the interior of ABC.2.D is EQUIDISTANT from BA and
BC.3.DA = DC4.DA _BA_, __DC_ BC5. DAB = 90°DCB = 90°6.__________7.BD ≅ BD8.__________9. ABD ≅ CBD10.BD bisects ABC and
point D is on the bisector of ABC
A
B
C
D
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.
Given: D is in the interior of ABC and is equidistant from BA and BC.Prove: D lies on the angle bisector of ABC.
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Statements:1. D is in the interior of ABC.2. D is EQUIDISTANT from BA and
BC.3. DA = DC4. DA _BA_, __DC_ BC
5. DAB and DCB are rt. s
6. DAB = 90°DCB = 90°7. BD ≅ BD8. __________9. ABD ≅ CBD10. BD bisects ABC and point D
is on the bisector of ABC
A
B
C
D
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
Given: D is in the interior of ABC and is equidistant from BA and BC.Prove: D lies on the angle bisector of ABC.
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Statements:1. D is in the interior of ABC.2. D is EQUIDISTANT from BA and
BC.3. DA = DC4. DA _BA_, __DC_ BC
5. DAB and DCB are rt. s
6. DAB = 90°DCB = 90°7. BD ≅ BD8. __________9. ABD ≅ CBD10. BD bisects ABC and point D
is on the bisector of ABC
A
B
C
D
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.
Given: D is in the interior of ABC and is equidistant from BA and BC.Prove: D lies on the angle bisector of ABC.
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Statements:1. D is in the interior of ABC.2. D is EQUIDISTANT from BA and
BC.3. DA = DC4. DA _BA_, __DC_ BC
5. DAB and DCB are rt. s
6. DAB = 90°DCB = 90°7. BD ≅ BD8. ∆ABD ≅ ∆CBD9. ABD ≅ CBD10. BD bisects ABC and point D
is on the bisector of ABC
A
B
C
D
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.
Given: D is in the interior of ABC and is equidistant from BA and BC.Prove: D lies on the angle bisector of ABC.
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Statements:1. D is in the interior of ABC.2. D is EQUIDISTANT from BA and
BC.3. DA = DC4. DA _BA_, __DC_ BC
5. DAB and DCB are rt. s
6. DAB = 90°DCB = 90°7. BD ≅ BD8. ∆ABD ≅ ∆CBD9. ABD ≅ CBD10. BD bisects ABC and point D
is on the bisector of ABC
A
B
C
D
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
Given: D is in the interior of ABC and is equidistant from BA and BC.Prove: D lies on the angle bisector of ABC.
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Statements:1. D is in the interior of ABC.2. D is EQUIDISTANT from BA and
BC.3. DA = DC4. DA _BA_, __DC_ BC
5. DAB and DCB are rt. s
6. DAB = 90°DCB = 90°7. BD ≅ BD8. ∆ABD ≅ ∆CBD9. ABD ≅ CBD10. BD bisects ABC and point D
is on the bisector of ABC
A
B
C
D
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.
Given: D is in the interior of ABC and is equidistant from BA and BC.Prove: D lies on the angle bisector of ABC.
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Assignment:
• You’ll do page 251 #1-11 we’ll go over them then you’ll turn #’s 12-31; 41, 42, 55-63