Lesson 5-1

Bisectors, Medians and Altitudes

5-Minute Check on Chapter 45-Minute Check on Chapter 45-Minute Check on Chapter 45-Minute Check on Chapter 4 Transparency 5-1

Refer to the figure.1. Classify the triangle as scalene, isosceles, or equilateral.

2. Find x if mA = 10x + 15, mB = 8x – 18, andmC = 12x + 3.

3. Name the corresponding congruent angles if RST UVW.

4. Name the corresponding congruent sides if LMN OPQ.

5. Find y if DEF is an equilateral triangle and mF = 8y + 4.

6. What is the slope of a line that contains (–2, 5) and (1, 3)?Standardized Test Practice:

A CB D–3/2–2/3 2/3 3/2

5-Minute Check on Chapter 45-Minute Check on Chapter 45-Minute Check on Chapter 45-Minute Check on Chapter 4 Transparency 5-1

Refer to the figure.1. Classify the triangle as scalene, isosceles, or equilateral.

isosceles

2. Find x if mA = 10x + 15, mB = 8x – 18, andmC = 12x + 3. 6

3. Name the corresponding congruent angles if RST UVW. R U; S V; T W

4. Name the corresponding congruent sides if LMN OPQ.LM OP; MN PQ; LN OQ

5. Find y if DEF is an equilateral triangle and mF = 8y + 4. 7

6. What is the slope of a line that contains (–2, 5) and (1, 3)?Standardized Test Practice:

A CB D–3/2–2/3 2/3 3/2

Objectives

• Identify and use perpendicular bisectors and angle bisectors in triangles

• Identify and use medians and altitudes in triangles

Vocabulary• Concurrent lines – three or more lines that intersect at

a common point

• Point of concurrency – the intersection point of three or more lines

• Perpendicular bisector – passes through the midpoint of the segment (triangle side) and is perpendicular to the segment

• Median – segment whose endpoints are a vertex of a triangle and the midpoint of the side opposite the vertex

• Altitude – a segment from a vertex to the line containing the opposite side and perpendicular to the line containing that side

Vocabulary• Circumcenter – the point of concurrency of the

perpendicular bisectors of a triangle; the center of the largest circle that contains the triangle’s vertices

• Centroid – the point of concurrency for the medians of a triangle; point of balance for any triangle

• Incenter – the point of concurrency for the angle bisectors of a triangle; center of the largest circle that can be drawn inside the triangle

• Orthocenter – intersection point of the altitudes of a triangle; no special significance

Theorems• Theorem 5.1 – Any point on the perpendicular bisector of a

segment is equidistant from the endpoints of the segment.• Theorem 5.2 – Any point equidistant from the endpoints of

the segments lies on the perpendicular bisector of a segment.

• Theorem 5.3, Circumcenter Theorem – The circumcenter of a triangle is equidistant from the vertices of the triangle.

• Theorem 5.4 – Any point on the angle bisector is equidistant from the sides of the triangle.

• Theorem 5.5 – Any point equidistant from the sides of an angle lies on the angle bisector.

• Theorem 5.6, Incenter Theorem – The incenter of a triangle is equidistant from each side of the triangle.

• Theorem 5.7, Centroid Theorem – The centroid of a triangle is located two thirds of the distance from a vertex to the midpoint of the side opposite the vertex on a median.

Triangles – Perpendicular Bisectors

C

Circumcenter

Note: from Circumcenter Theorem: AP = BP = CP

Midpoint of AB

Midpoint of BC

Midpoint of ACZ

Y

XP

B

A

Circumcenter is equidistant from the vertices

Triangles – Angle Bisectors

B

C

A

Incenter

Note: from Incenter Theorem: QX = QY = QZ

Z

X

Y

Q

Incenter is equidistant from the sides

Triangles – Medians

C

Midpoint of AB

Midpoint of BC

Midpoint of AC

Centroid

Medianfrom B

M

Z

Note: from Centroid theorem BM = 2/3 BZ

Y

X

Centroid is the point of balance in any triangle

B

A

Triangles – Altitudes

C

X

Z

Y

Orthocenter

Altitudefrom B

Orthocenter has no special significance for us

B

ANote: Altitude is the shortest distance

from a vertex to the line opposite it

Special Segments in Triangles

Name TypePoint of

ConcurrencyCenter Special

QualityFrom / To

Perpendicular

bisector

Line, segment or

ray

Circumcenter

Equidistantfrom vertices

Nonemidpoint of

segment

Angle bisector

Line, segment or

ray

IncenterEquidistantfrom sides

Vertexnone

Median segment CentroidCenter ofGravity

Vertexmidpoint of

segment

Altitude segment Orthocenter none Vertexnone

Location of Point of Concurrency

Name Point of Concurrency Triangle ClassificationAcute Right Obtuse

Perpendicular bisector Circumcenter Inside hypotenuse Outside

Angle bisector Incenter Inside Inside Inside

Median Centroid Inside Inside Inside

Altitude Orthocenter Inside Vertex - 90 Outside

Given:

Find:

Proof:

Statements Reasons

1. 1. Given

2. 2. Angle Sum Theorem3. 3. Substitution4. 4. Subtraction Property 5. 5. Definition of angle

bisector6. 6. Angle Sum Theorem7. 7. Substitution8. 8. Subtraction Property

mDGE

Find:

Given:

.

Proof:

Statements. Reasons

1. Given

2. Angle Sum Theorem3. Substitution4. Subtraction Property 5. Definition of angle bisector6. Angle Sum Theorem7. Substitution8. Subtraction Property

1.

2.3.4.5.

6.7.8.

1. Given

mADC

ALGEBRA Points U, V, and W are the midpoints of respectively. Find a, b, and c.

Find a.

Segment Addition Postulate

Centroid Theorem

Substitution

Multiply each side by 3 and simplify.

Subtract 14.8 from each side.

Divide each side by 4.

Find b.

Segment Addition Postulate

Centroid Theorem

Substitution

Multiply each side by 3 and simplify.

Subtract 6b from each side.

Divide each side by 3.

Subtract 6 from each side.

Find c.

Segment Addition Postulate

Centroid Theorem

Substitution

Multiply each side by 3 and simplify.

Subtract 30.4 from each side.

Divide each side by 10.

Answer:

ALGEBRA Points T, H, and G are the midpoints of respectively. Find w, x, and y.

Answer:

Summary & Homework

• Summary:– Perpendicular bisectors, angle bisectors, medians

and altitudes of a triangle are all special segments in triangles

– Perpendiculars and altitudes form right angles– Perpendiculars and medians go to midpoints– Angle bisector cuts angle in half

• Homework: – Day 1: pg 245: 46-49– Day 2: pg 245: 51-54