Lesson 5-1

Bisectors in Triangles

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 in triangles

• Identify and use angle bisectors 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

• 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

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.

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

Perpendicular Bisector Theorems

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.

Example 1B

B. Find XY.

Answer: 6

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.

Perpendicular Bisector Theorems

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.

Triangles – Angle Bisectors

B

C

A

Incenter

Note: from Incenter Theorem: QX = QY = QZ

Z

X

Y

Q

Incenter is equidistant from the sides

Angle Bisector Theorems

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

Example 3B

B. Find mWYZ.

Answer: mWYZ = 28°

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.

Angle Bisector Theorems

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

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:

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

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

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

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