5.1 Perpendiculars and Bisectors

GeometryMrs. SpitzFall 2004

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

Equidistant

• A point is equidistant from two points if its distance from each point is the same.

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

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 ∆APC and ∆BPC are congruent using the SAS Congruence Postulate. Then show that CA ≅ CB.

Given segment

perpendicular bisector

PA B

C

Statements:1. CP is perpendicular

bisector of AB.2. CP AB3. AP BP≅4. CP CP≅5. CPB ≅ CPA6. ∆APC ≅ ∆BPC7. CA ≅ CB

Reasons:1. Given

Given: CP is perpendicular to AB. AP BP≅

Prove: CA CB≅

perpendicular bisector

PA B

C

Statements:1. CP is perpendicular

bisector of AB.2. CP AB3. AP BP≅4. CP CP≅5. CPB ≅ CPA6. ∆APC ≅ ∆BPC7. CA ≅ CB

Reasons:1. Given2. Definition of

Perpendicular bisector

Given: CP is perpendicular to AB. AP BP≅

Prove: CA CB≅

perpendicular bisector

PA B

C

Statements:1. CP is perpendicular

bisector of AB.2. CP AB3. AP BP≅4. CP CP≅5. CPB ≅ CPA6. ∆APC ≅ ∆BPC7. CA ≅ CB

Reasons:1. Given2. Definition of

Perpendicular bisector

3. Given

Given: CP is perpendicular to AB. AP BP≅

Prove: CA CB≅

perpendicular bisector

PA B

C

Statements:1. CP is perpendicular

bisector of AB.2. CP AB3. AP BP≅4. CP CP≅5. CPB ≅ CPA6. ∆APC ≅ ∆BPC7. CA ≅ CB

Reasons:1. Given2. Definition of

Perpendicular bisector3. Given4. Reflexive Prop.

Congruence.

Given: CP is perpendicular to AB. AP BP≅

Prove: CA CB≅

perpendicular bisector

PA B

C

Statements:1. CP is perpendicular

bisector of AB.2. CP AB3. AP BP≅4. CP CP≅5. 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. AP BP≅

Prove: CA CB≅

perpendicular bisector

PA B

C

Statements:1. CP is perpendicular

bisector of AB.2. CP AB3. AP BP≅4. CP CP≅5. 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. AP BP≅

Prove: CA CB≅

perpendicular bisector

PA B

C

Statements:1. CP is perpendicular

bisector of AB.2. CP AB3. AP BP≅4. CP CP≅5. CPB ≅ CPA6. ∆APC ≅ ∆BPC7. CA ≅ CB

Reasons:1. Given2. Definition of

Perpendicular bisector3. Given4. Reflexive Prop.

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

Given: CP is perpendicular to AB. AP BP≅

Prove: CA CB≅

perpendicular bisector

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

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

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

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

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 Theorem 5.3

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