5.1 Perpendiculars and Bisectors Geometry Mrs. Spitz Fall 2004

5.1 Perpendiculars and Bisectors

GeometryMrs. SpitzFall 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

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. 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.

CP is a bisector of AB

Perpendicular Bisector Construction – pg. 264

1. If you cannot follow these directions, to go page 264 for the instructions with pictures.

2. Draw a line. Any line about the middle of the page.3. Place compass point at P. Draw an arc that intersects

line m twice. Label the intersections as A and B.4. 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.

5. Use a straightedge to draw CP. This line is perpendicular to line m and passes through P.

6. 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.

If CP is the perpendicular bisector of AB, then CA = CB.

Given segment


PA B

C

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.

• Exercise 28 asks you to write a two-column proof of Theorem 5.1 using this plan. (This is part of your homework)

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

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 mBAD = mCAD, then DB = DC

B

A

C

D

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 mBAD = mCAD.

B

A

C

D

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

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

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?

B

A

O

C D

G KHLM 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.

B

A

O

C D

G KHLM 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

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.


Documents

5.1 Perpendiculars and Bisectors Geometry Mrs. Spitz Fall 2004