5.1 midsegments of triangles geometry mrs. spitz fall 2004

5.1 midsegments of triangles

GeometryMrs. SpitzFall 2004

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.

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.

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.

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

Example 2

• In triangle DEF, A, B, and C are midpoints. Name pairs of parallel segments.

• 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=

=

||

Real world connection

• CD is a new bridge being built over a lake as shown. Find the length of the bridge.

Assignment:

• Page 246 #1-20 we’ll go over them then you’ll turn in tomorrow #’s 21-36; 47-55

5.2 Bisectors in Triangles

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.

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

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.

Perpendicular bisector theorem

• If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.

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


PA B

C

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.

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

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


PA B

C

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



Reasons:1. Given2. Definition of

Perpendicular bisector



PA B

C




Perpendicular bisector

3. Given



PA B

C




Perpendicular bisector3. Given4. Reflexive Prop.

Congruence.



PA B

C





Congruence.5. Definition right angle



PA B

C





Congruence.5. Definition right angle6. SAS Congruence



PA B

C





Congruence.5. Definition right angle6. SAS Congruence7. CPCTC



PA B

C

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


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


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

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

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

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

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

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

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.

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


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


A

B

C

D

Reasons:1. Given

2. Given

3. Def. Equidistant


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


A

B

C

D

Reasons:

1. Given

2. Given

3. Def. Equidistant

4. Def. Distance from point to line.


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


A

B

C

D

Reasons:1. Given

2. Given

3. Def. Equidistant


5. If 2 lines are , then they form 4 rt. s.


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


A

B

C

D

Reasons:

1. Given

2. Given

3. Def. Equidistant



6. Def. of a Right Angle





6. DAB = 90°DCB = 90°7. BD ≅ BD8. __________9. ABD ≅ CBD10. BD bisects ABC and point D


A

B

C

D

Reasons:

1. Given

2. Given

3. Def. Equidistant




7. Reflexive Property of Cong.





6. DAB = 90°DCB = 90°7. BD ≅ BD8. ∆ABD ≅ ∆CBD9. ABD ≅ CBD10. BD bisects ABC and point D


A

B

C

D

Reasons:

1. Given

2. Given

3. Def. Equidistant





8. HL Congruence Thm.







A

B

C

D

Reasons:

1. Given

2. Given

3. Def. Equidistant






9. CPCTC







A

B

C

D

Reasons:

1. Given

2. Given

3. Def. Equidistant






9. CPCTC

10. Angle Bisector Thm.


Assignment:

• You’ll do page 251 #1-11 we’ll go over them then you’ll turn #’s 12-31; 41, 42, 55-63

5.1 midsegments of triangles geometry mrs. spitz fall 2004

Documents

segment bisector

perpendicular bisector

midsegmenta segment

use properties of midsegments

proof of theorem

sidesmidsegment theorem

midsegments of trianglesturn

triangles objectives