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# 5.1 Perpendiculars and Bisectors - PowerPoint PPT Presentation

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.

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### 5.1 Perpendiculars and Bisectors

Geometry

Mrs. Spitz

Fall 2004

• Use properties of perpendicular bisectors

• Use properties of angle bisectors to identify equal distances such as the lengths of beams in a room truss.

• pp. 267-269 #1-26, 28, 32-35 all – Quiz after 5.3

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.

Use Properties of perpendicular bisectors

CP is a  bisector of AB

Perpendicular Bisector Construction – pg. 264 intersects a segment at its midpoint. A segment, ray, line, or plane that is perpendicular to a segment at its midpoint is called a

• 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 intersects a segment at its midpoint. A segment, ray, line, or plane that is perpendicular to a segment at its midpoint is called a

• 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 intersects a segment at its midpoint. A segment, ray, line, or plane that is perpendicular to a segment at its midpoint is called a

• 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.

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.

Theorem 5.1 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.

Theorem 5.2: Converse of the Perpendicular Bisector Theorem

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)

Plan for Proof of Theorem 5.1

Statements: given that CP is the perpendicular bisector of AB. Show that right triangles

CP is perpendicular bisector of AB.

CP  AB

AP ≅ BP

CP ≅ CP

CPB ≅ CPA

∆APC ≅ ∆BPC

CA ≅ CB

Reasons:

Given

Given: CP is perpendicular to AB.

Prove: CA≅CB

Statements: given that CP is the perpendicular bisector of AB. Show that right triangles

CP is perpendicular bisector of AB.

CP  AB

AP ≅ BP

CP ≅ CP

CPB ≅ CPA

∆APC ≅ ∆BPC

CA ≅ CB

Reasons:

Given

Definition of Perpendicular bisector

Given: CP is perpendicular to AB.

Prove: CA≅CB

Statements: given that CP is the perpendicular bisector of AB. Show that right triangles

CP is perpendicular bisector of AB.

CP  AB

AP ≅ BP

CP ≅ CP

CPB ≅ CPA

∆APC ≅ ∆BPC

CA ≅ CB

Reasons:

Given

Definition of Perpendicular bisector

Given

Given: CP is perpendicular to AB.

Prove: CA≅CB

Statements: given that CP is the perpendicular bisector of AB. Show that right triangles

CP is perpendicular bisector of AB.

CP  AB

AP ≅ BP

CP ≅ CP

CPB ≅ CPA

∆APC ≅ ∆BPC

CA ≅ CB

Reasons:

Given

Definition of Perpendicular bisector

Given

Reflexive Prop. Congruence.

Given: CP is perpendicular to AB.

Prove: CA≅CB

Statements: given that CP is the perpendicular bisector of AB. Show that right triangles

CP is perpendicular bisector of AB.

CP  AB

AP ≅ BP

CP ≅ CP

CPB ≅ CPA

∆APC ≅ ∆BPC

CA ≅ CB

Reasons:

Given

Definition of Perpendicular bisector

Given

Reflexive Prop. Congruence.

Definition right angle

Given: CP is perpendicular to AB.

Prove: CA≅CB

Statements: given that CP is the perpendicular bisector of AB. Show that right triangles

CP is perpendicular bisector of AB.

CP  AB

AP ≅ BP

CP ≅ CP

CPB ≅ CPA

∆APC ≅ ∆BPC

CA ≅ CB

Reasons:

Given

Definition of Perpendicular bisector

Given

Reflexive Prop. Congruence.

Definition right angle

SAS Congruence

Given: CP is perpendicular to AB.

Prove: CA≅CB

Statements: given that CP is the perpendicular bisector of AB. Show that right triangles

CP is perpendicular bisector of AB.

CP  AB

AP ≅ BP

CP ≅ CP

CPB ≅ CPA

∆APC ≅ ∆BPC

CA ≅ CB

Reasons:

Given

Definition of Perpendicular bisector

Given

Reflexive Prop. Congruence.

Definition right angle

SAS Congruence

CPCTC

Given: CP is perpendicular to AB.

Prove: CA≅CB

In the diagram MN is the perpendicular bisector of ST. given that CP is the perpendicular bisector of AB. Show that right triangles

What segment lengths in the diagram are equal?

Explain why Q is on MN.

Ex. 1 Using Perpendicular Bisectors

What segment lengths in the diagram are equal? given that CP is the perpendicular bisector of AB. Show that right triangles

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.

Ex. 1 Using Perpendicular Bisectors

Explain why Q is on MN. given that CP is the perpendicular bisector of AB. Show that right triangles

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.

Ex. 1 Using Perpendicular 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.

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.

Using Properties of Angle Bisectors

If a point is on the bisector of an angle, then it is equidistant from the two sides of the angle.

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.

Theorem 5.3 Angle Bisector Theorem

Given: D is on the bisector of from the sides of the angle, then it lies on the bisector of the angle.BAC. DB AB, DC  AC.

Prove: DB = DC

Plan for Proof: Prove that ∆ADB ≅ ∆ADC. Then conclude that DB ≅DC, so DB = DC.

Ex. 2: Proof of Theorem 4.3

By definition of an angle bisector, from the sides of the angle, then it lies on the bisector of the angle.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.

Paragraph Proof

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?

Ex. 3: Using Angle Bisectors

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.

SOLUTION:

Given: D is in the interior of perpendicular segments to the sides of ABC and is equidistant from BA and BC.

Prove: D lies on the angle bisector of ABC.

32. Developing Proof

Statements: perpendicular segments to the sides of

D is in the interior of ABC.

D is ___?_ from BA and BC.

____ = ____

DA  ____, ____  BC

__________

__________

BD ≅ BD

__________

ABD ≅ CBD

BD bisects ABC and point D is on the bisector of ABC

Given: D is in the interior of ABC and is equidistant from BA and BC.

Prove: D lies on the angle bisector of ABC.

Reasons:

• Given

Statements: perpendicular segments to the sides of

D is in the interior of ABC.

D is EQUIDISTANT from BA and BC.

____ = ____

DA  ____, ____  BC

__________

__________

BD ≅ BD

__________

ABD ≅ CBD

BD bisects ABC and point D is on the bisector of ABC

Given: D is in the interior of ABC and is equidistant from BA and BC.

Prove: D lies on the angle bisector of ABC.

Reasons:

• Given

• Given

Statements: perpendicular segments to the sides of

D is in the interior of ABC.

D is EQUIDISTANT from BA and BC.

DA = DC

DA  ____, ____  BC

__________

__________

BD ≅ BD

__________

ABD ≅ CBD

BD bisects ABC and point D is on the bisector of ABC

Given: D is in the interior of ABC and is equidistant from BA and BC.

Prove: D lies on the angle bisector of ABC.

Reasons:

• Given

• Given

• Def. Equidistant

Statements: perpendicular segments to the sides of

D is in the interior of ABC.

D is EQUIDISTANT from BA and BC.

DA = DC

DA  _BA_, __DC_  BC

__________

__________

BD ≅ BD

__________

ABD ≅ CBD

BD bisects ABC and point D is on the bisector of ABC

Given: D is in the interior of ABC and is equidistant from BA and BC.

Prove: D lies on the angle bisector of ABC.

Reasons:

• Given

• Given

• Def. Equidistant

• Def. Distance from point to line.

Statements: perpendicular segments to the sides of

D is in the interior of ABC.

D is EQUIDISTANT from BA and BC.

DA = DC

DA  _BA_, __DC_  BC

DAB = 90°DCB = 90°

__________

BD ≅ BD

__________

ABD ≅ CBD

BD bisects ABC and point D is on the bisector of ABC

Given: D is in the interior of ABC and is equidistant from BA and BC.

Prove: D lies on the angle bisector of ABC.

Reasons:

• Given

• Given

• Def. Equidistant

• Def. Distance from point to line.

• If 2 lines are , then they form 4 rt. s.

Statements: perpendicular segments to the sides of

D is in the interior of ABC.

D is EQUIDISTANT from BA and BC.

DA = DC

DA  _BA_, __DC_  BC

DAB and DCB are rt. s

DAB = 90°DCB = 90°

BD ≅ BD

__________

ABD ≅ CBD

BD bisects ABC and point D is on the bisector of ABC

Given: D is in the interior of ABC and is equidistant from BA and BC.

Prove: D lies on the angle bisector of ABC.

Reasons:

• Given

• Given

• Def. Equidistant

• Def. Distance from point to line.

• If 2 lines are , then they form 4 rt. s.

• Def. of a Right Angle

Statements: perpendicular segments to the sides of

D is in the interior of ABC.

D is EQUIDISTANT from BA and BC.

DA = DC

DA  _BA_, __DC_  BC

DAB and DCB are rt. s

DAB = 90°DCB = 90°

BD ≅ BD

__________

ABD ≅ CBD

BD bisects ABC and point D is on the bisector of ABC

Given: D is in the interior of ABC and is equidistant from BA and BC.

Prove: D lies on the angle bisector of ABC.

Reasons:

• Given

• Given

• Def. Equidistant

• Def. Distance from point to line.

• If 2 lines are , then they form 4 rt. s.

• Def. of a Right Angle

• Reflexive Property of Cong.

Statements: perpendicular segments to the sides of

D is in the interior of ABC.

D is EQUIDISTANT from BA and BC.

DA = DC

DA  _BA_, __DC_  BC

DAB and DCB are rt. s

DAB = 90°DCB = 90°

BD ≅ BD

∆ABD ≅ ∆CBD

ABD ≅ CBD

BD bisects ABC and point D is on the bisector of ABC

Given: D is in the interior of ABC and is equidistant from BA and BC.

Prove: D lies on the angle bisector of ABC.

Reasons:

• Given

• Given

• Def. Equidistant

• Def. Distance from point to line.

• If 2 lines are , then they form 4 rt. s.

• Def. of a Right Angle

• Reflexive Property of Cong.

• HL Congruence Thm.

Statements: perpendicular segments to the sides of

D is in the interior of ABC.

D is EQUIDISTANT from BA and BC.

DA = DC

DA  _BA_, __DC_  BC

DAB and DCB are rt. s

DAB = 90°DCB = 90°

BD ≅ BD

∆ABD ≅ ∆CBD

ABD ≅ CBD

BD bisects ABC and point D is on the bisector of ABC

Given: D is in the interior of ABC and is equidistant from BA and BC.

Prove: D lies on the angle bisector of ABC.

Reasons:

• Given

• Given

• Def. Equidistant

• Def. Distance from point to line.

• If 2 lines are , then they form 4 rt. s.

• Def. of a Right Angle

• Reflexive Property of Cong.

• HL Congruence Thm.

• CPCTC

Statements: perpendicular segments to the sides of

D is in the interior of ABC.

D is EQUIDISTANT from BA and BC.

DA = DC

DA  _BA_, __DC_  BC

DAB and DCB are rt. s

DAB = 90°DCB = 90°

BD ≅ BD

∆ABD ≅ ∆CBD

ABD ≅ CBD

BD bisects ABC and point D is on the bisector of ABC

Given: D is in the interior of ABC and is equidistant from BA and BC.

Prove: D lies on the angle bisector of ABC.

Reasons:

• Given

• Given

• Def. Equidistant

• Def. Distance from point to line.

• If 2 lines are , then they form 4 rt. s.

• Def. of a Right Angle

• Reflexive Property of Cong.

• HL Congruence Thm.

• CPCTC

• Angle Bisector Thm.