71
5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Embed Size (px)

Citation preview

Page 1: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

5.1 Perpendiculars and Bisectors

Day 1 Part 1

CA Standard 16.0

Page 2: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Warmup

Simplify. 1. 6x + 11y – 4x + y 2. -5m + 3q + 4m – q 3. -3q – 4t – 5t – 2p 4. 9x – 22y + 18x – 3y 5. 5x2 + 2xy – 7x2 + xy

Page 3: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

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.

A P B

C

Page 4: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Converse of the Perpendicular Bisector Theorem

Page 5: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

In the diagram shown, MN is the perpendicular bisector of ST.

What segment lengths in the diagram are equal?

Explain why Q is on MN

T

M N

S

Q

12

12

Page 6: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

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 m<BAD = m<CAD, then DB = DC.

A

B

C

D

Page 7: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Converse of the Angle Bisector Theorem

Page 8: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Use the diagram to answer the following. In the diagram, F is on the bisector of

< DAE. If m<BAF = 50, then m<CAF = ____ If FC = 10, then FB = ____ Is triangle ABF congruent to triangle ACF?

Explain. A

B

D

G

F

C

E

Page 9: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0
Page 10: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

5.2 Bisectors of a Triangle

Day 1 Part 2

CA Standards 16.0, 21.0

Page 11: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

In the figure, YW bisects <XYZ.

m<XYZ = 6x + 2, m<ZYW = 8x – 6.

Solve for x and find m<XYZ.

Y

X

W

Z

Page 12: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Concurrency of Perpendicular Bisectors of a

Triangle The perpendicular bisectors of a triangle

intersect at a point

that is equidistant from

the vertices of the

triangle.

PA = PB = PCA

B

C

P

P is also called the circumcenter of the triangle.

Page 13: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Use the diagram shown.

E is the circumcenter of Δ ABC. DA = ___ BF = ___ <EFC = ___

A

D

B

E

CF

Page 14: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Definitions

Concurrent lines: three or more lines intersect in the same point.

Point of concurrency: the point of intersection of the lines.

Page 15: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Concurrency of Angle Bisectors of a Triangle

The angle bisectors of a triangle intersect at a point that is equidistant from the vertices of the triangle.

PD = PE = PF

A

B

C

P

E

D F

Page 16: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

The point of concurrency can be inside the triangle, on the triangle, or outside of the triangle. Acute Triangle: inside Right Triangle: on Obtuse Triangle: outside

Page 17: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Example

Which segments

are congruent?

M

P N

RQ

L

S

Page 18: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Use the diagram shown.

E is the circumcenter of Δ ABC. DA = ___ BF = ___ <EFC = ___

A

D

B

E

CF

Page 19: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Mini quiz on definitions…

The _____________ of the angle bisectors is called the incenter of the triangle.

If three or more lines intersect at the same point, the lines are ________.

The point of concurrency of the perpendicular bisectors of a triangle is called ____________________.

Point of concurrency

Concurrent

Circumcenter of the triangle

Page 20: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Construction

Pg. 268 # 14, 15 Pg. 275 # 5 – 9

Page 21: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Pg. 269 # 21 – 29, 32 Pg. 275 # 10 - 22

Page 22: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

5.3 Medians and Altitudes of a Triangle

Day 2 Part 1

CA Standards 16.0

Page 23: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Warmup

Find BD.

12 12

15

C

D

B A

Page 24: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

AC = ___ m<DCB = ___ m<B = ___

A

B

C20

L

D

55

20

55

35

Page 25: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Median of a triangle.

Median of a triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side.

Median

Page 26: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

The three medians of a triangle are concurrent. The point of concurrency is called the centroid of the triangle.

•P

Centroid

Page 27: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Concurrency of Medians of a Triangle

The medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side.

AP = 2/3 AD

BP = 2/3 BF

CP = 2/3 CE

A B

C

DF

E

P

Page 28: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

P is the centroid of ∆QRS shown. Find RT and RP when PT = 5.

R

SQ T

P

Page 29: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Sketch ∆JKL with J(7,10), K(5,2), L(3,6). Find the coordinates of the centroid of ∆JKL.

Page 30: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Altitude of a triangle

An altitude of a triangle is the perpendicular segment from a vertex to the opposite side or to the line that contains the opposite side.

Altitude

Page 31: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Every triangle has three altitudes.

The lines containing the altitudes are concurrent and intersect at a point: orthocenter of the triangle.

Where is the orthocenter located in each type of triangle? Acute triangle Right triangle Obtuse triangle

Page 32: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Use the diagram shown and the given information to decide in each case whether EG is a

perpendicular bisector, an angle bisector, a median, or an altitude of Δ DEF.

1. DG = FG

2. EG DF

3. m<DEG = m<FEGT

E

D G F

Median

Altitude

Angle bisector

Page 33: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

The angle bisectors of Δ ABC meet at point D. Find DE.

A

B

C

D

E F

G

LL19

28

19

Page 34: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0
Page 35: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

5.4 Midsegment Theorem

Day 2 Part 2

CA Standards 17.0

Page 36: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Review

Given PQ = 14, SU = 6, and QU = 3, find the perimeter of Δ STU.

Q

U

R

S

T P

Page 37: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Midsegment Theorem

The segment connecting the midpoint of two sides of a triangle is parallel to the third side and is half as long.

DE ll AB and DE = ½ AB

A B

C

D E>

>

Page 38: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0
Page 39: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

UW and VW are midsegment of Δ RST. Find UW and RT.

R

S

V

U

W

T812 6

16

Page 40: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

GH, HJ and JG are midsegments of Δ EDF.

1. JH ll ___

2. EF = ___

3. DF = ___

4. ___ ll DE

5. GH = ___

6. JH = ___

D

G

F

H

EJ

24

810.6

DF

21.2

16

GH

12

8

Page 41: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0
Page 42: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Given the midpoints of a triangle are (7,4), (5,6) and (8,7), find the coordinates of the vertices.

Page 43: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Pg. 282 # 3 – 12 Pg. 283 # 17 – 20 Pg. 290 # 3 – 22, 26 – 29

Page 44: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

5.5 Inequalities in One Triangle

Day 3 Part 1

CA Standards 6.0, 13.0

Page 45: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Warmup

Solve the inequality. 1. -x + 2 > 7 2. c – 18 < 10 3. -5 + m < 21 x – 5 > 4 z + 6 > -2

Page 46: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Theorems

If one side of a triangle is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side.

m<B > m<CA

B

C

3 7

Page 47: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

List the angles in order from greatest to least.

A

B C

27

23

18

Page 48: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

If one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle.

D

E

F

60°

40°

EF > DF

Page 49: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Write the measurements of the triangles in order from least to greatest.

J

H

G

100°

35°

45°

Q

R

P

5

6

7

JG, HJ, HG m<R, m<Q, m<P

Page 50: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

List the sides in order from longest to shortest.

G E

F

45° 65°

Page 51: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Exterior Angle Inequality Theorem

The measure of an exterior angle of a triangle is greater than the measure of either of the two nonadjacent (not next to) interior angles.

A

B C1

m< 1 > m<A

m<1 > m<B

Page 52: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Constructing a Triangle

Construct a triangle with the given group of side length, if possible. 4 in, 4 in, 4 in

2 in, 4 in, 6 in

3 in, 4 in, 5 in

Page 53: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Triangle Inequality Theorem

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

AB + BC > AC

AC + BC > AB

AB + AC > BCA

B

C

Page 54: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Use the diagram to solve the inequality

AB + BC > AC.

A

B

C

6x + 3

4x + 5

3x + 2

Page 55: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Two sides of a triangle have lengths 4 and 14. Describe the possible length of the third side.

Two sides of a triangle have lengths 10 and 15. Describe the possible length of the third side.

Page 56: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

5.6 Indirect Proof and Inequalities in Two Triangles

Day 3 Part 2

CA Standards 2.0, 6.0, 13.0

Page 57: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Review

1. What is the sum of x and y?

2. Which measure is greater, x ° or y °?

x

y21

20

Page 58: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Hinge Theorem

If two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is larger than the included angle of the second, then the third side of the first is longer than the third side of the second.

R

S

T

W

V

X100° 80°

RT > WX

Page 59: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Use Hinge theorem to complete the blank with <, >, or =.

1. RS __ TU 2. m<1 ___ m< 2

3. XY __ ZY

R

S

T

U110°

130°

1

2

15 13

< >

X

Y

Z

41°

38°

<

Page 60: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Converse of the Hinge Theorem

Page 61: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

What is the largest angle that is part of a triangle?

9910

44

W

XY Z

Page 62: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Review List the sides in order from shortest to

longest.L

M

N

75°

74°

A

B

C50°

49°

75° + 74°+ m<N = 180°149° + m<N = 180°m<N = 31°

31°

LM, LN, MN

49 ° + 50 ° + m<B = 180 °99 ° + m<B = 180 °m<B = 81 °

81 °

BC, AB, AC

Page 63: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Review

Solve for a

2a + 7

a + 19

Page 64: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Extra Credit!!

Use the diagram to solve.

1. Find the value of x.

2. Find m<B

3. Find m<C

4. Find m<BAC

D

A

BC

3x°

(x+13)°(x+19)°

Page 65: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Pg. 298 # 6 – 25, 34 Pg. 305 # 3 – 23, 26, 27

Page 66: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Ch 5 Review

Day 4 Part 1

Page 67: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Warmup

Page 68: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Review

Find the value of x.

40

x

2x+6

Page 69: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Review

Page 70: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Ch 5 Culminating Task

Page 71: 5.1 Perpendiculars and Bisectors Day 1 Part 1 CA Standard 16.0

Ch 5 Test

Day 4 Part 2