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Lesson 5-1. Bisectors in Triangles. Transparency 5-1. 5-Minute Check on Chapter 4. Refer to the figure. 1. Classify the triangle as scalene, isosceles, or equilateral. 2. Find x if m  A = 10 x + 15, m  B = 8 x – 18, and m  C = 12 x + 3. - PowerPoint PPT Presentation
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Lesson 5-1 Bisectors in Triangles
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Page 1: Lesson 5-1

Lesson 5-1

Bisectors in Triangles

Page 2: Lesson 5-1

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

Page 3: Lesson 5-1

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

Page 4: Lesson 5-1

Objectives

• Identify and use perpendicular bisectors in triangles

• Identify and use angle bisectors in triangles

Page 5: Lesson 5-1

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

• Circumcenter – the point of concurrency of the perpendicular bisectors of a triangle; the center of the largest circle that contains the triangle’s vertices

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

Page 6: Lesson 5-1

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.

Page 7: Lesson 5-1

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

Page 8: Lesson 5-1

Perpendicular Bisector Theorems

Page 9: Lesson 5-1

Example 1A

A. Find BC.

Answer: 8.5

BC = AC Perpendicular Bisector Theorem

BC = 8.5 Substitution

From the information in the diagram, we know that line CD is the perpendicular bisector of line segment AB.

Page 10: Lesson 5-1

Example 1B

B. Find XY.

Answer: 6

Page 11: Lesson 5-1

Example 1C

C. Find PQ.

Answer: 7

PQ = RQ Perpendicular Bisector Theorem

3x + 1 = 5x – 3 Substitution

1 = 2x – 3 Subtract 3x from each side.

4 = 2x Add 3 to each side.

2 = x Divide each side by 2.

So, PQ = 3(2) + 1 = 7.

From the information in the diagram, we know that line QS is the perpendicular bisector of line segment PR.

Page 12: Lesson 5-1

Perpendicular Bisector Theorems

Page 13: Lesson 5-1

Example 2

• GARDEN A triangular-shaped garden is shown. Can a fountain be placed at the circumcenter and still be inside the garden?

Answer: No, the circumcenter of an obtuse triangle is in the exterior of the triangle.

By the Circumcenter Theorem, a point equidistant from three points is found by using the perpendicular bisectors of the triangle formed by those points.

Copy ΔXYZ, and use a ruler and protractor to draw the perpendicular bisectors. The location for the fountain is C, the circumcenter of ΔXYZ, which lies in the exterior of the triangle.

Page 14: Lesson 5-1

Triangles – Angle Bisectors

B

C

A

Incenter

Note: from Incenter Theorem: QX = QY = QZ

Z

X

Y

Q

Incenter is equidistant from the sides

Page 15: Lesson 5-1

Angle Bisector Theorems

Page 16: Lesson 5-1

Example 3A

A. Find DB.

Answer: 5

From the information in the diagram, we know that ray AD is the angle bisector of BAC.

DB = DC Angle Bisector Theorem

DB = 5 Substitution

Page 17: Lesson 5-1

Example 3B

B. Find mWYZ.

Answer: mWYZ = 28°

Page 18: Lesson 5-1

Example 3C

C. Find QS.

Answer: QS = 4(3) – 1 = 11

QS = SR Angle Bisector Theorem

4x – 1 = 3x + 2 Substitution

x – 1 = 2 Subtract 3x from each side.

x = 3 Add 1 to each side.

Page 19: Lesson 5-1

Angle Bisector Theorems

Page 20: Lesson 5-1

Example 4AA. Find ST if S is the incenter of ΔMNP.

Answer: ST = SU = 6

By the Incenter Theorem, since S is equidistant from the sides of ΔMNP, ST = SU.

Find ST by using the Pythagorean Theorem.

a2 + b2 = c2 Pythagorean Theorem

82 + SU2 = 102 Substitution

64 + SU2 = 100 82 = 64, 102 = 100

SU2 = 36 subtract 36 from both sides

SU = 6 positive square root

Page 21: Lesson 5-1

Example 4B

B. Find mSPU if S is the incenter of ΔMNP.

Answer: mSPU = (1/2) mUPR = 31

mUPR + mRMT + mTNU = 180 Triangle Angle Sum Theorem

mUPR + 62 + 56 = 180 SubstitutionmUPR + 118 = 180 Simplify.

mUPR = 62 Subtract 118 from each side.

Since NR and MU are angle bisectors, we double the given half-angles to get the whole angles in:

Page 22: Lesson 5-1

Given:

Find: mDGE

Example 5

2x + 80 + 30 = 180 3 ’s in triangle sum to 180 2x + 110 = 180

2x = 70 subtracting 110 from both sides x = 35 dividing both sides by 2

x + mDGE + 30 = 180 3 ’s in triangle sum to 18035 + mDGE + 30 = 180 substitute x mDGE + 65 = 180 combine numbers mDGE = 115 subtract 65 from both sides

Page 23: Lesson 5-1

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

Page 24: Lesson 5-1

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

Page 25: Lesson 5-1

Summary & Homework

• Summary:– Perpendicular bisectors and angle bisectors of a

triangle are all special segments in triangles– Perpendiculars bisectors:

• form right angles• divide a segment in half – go through midpoints• equal distance from the vertexes of the triangle

– Angle bisector:• cuts angle in half• equal distance from the sides of the triangle

• Homework: – pg 327-31; 1-3, 5-7, 9-11, 26-30


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