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Journal Chapter 5 . Full screen to listen to music. TRIANGLES……. … TRIANGLES…. … AND… … MORE… TRIANGLES. Menu options ahead. Option #1. Perpendicular Bisector:. What is it? A line that bisects a segment and is perpendicular to that same segment. . THEOREM: .

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Journal Chapter 5

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Journal chapter 5

Journal

Chapter 5

Full screen to listen to music

TRIANGLES…….

TRIANGLES….

AND…

MORE…

TRIANGLES.

Menu options ahead.


Journal chapter 5

Option #1


Journal chapter 5

Perpendicular Bisector:

What is it?

A line that bisects a segment and is perpendicular to that same segment.

THEOREM:

* Any point that is on the perpendicular bisector is equidistant to both of the endpoints of the segment which is perpendicular to.

Converse:

*If the point is equidistant from both of the endpoints of the segment, then it is a perpendicular bisector.

Extra:

A perpendicular bisector is the locus of all points in a plane that are equidistant from the endpoints of a segment.


Journal chapter 5

7cm

Segment

Perpendicular bisector

Are both congruent.

7cm


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EXAMPLES

1. Given that AC= 25.5, BD=20, and BC= 25.5. Find AB.

AB= 51

2. Given that m is the perpendicular bisector of AB and BC= 30 . Find AC.

AC= 30

Given that m is the perpendicular bisector of AB , AC= 4a, and BC= 2a+26. Find BC.

2a+26=4a

-2a=-26

a=13 plug in both equations

BC= 52

a

d

m

c

b


Journal chapter 5

Option

#2


Journal chapter 5

Angle Bisectors

What is it?

A ray or any line that cuts an angle into two congruent angles.

Theorem :

* Any point that lies on the angle bisector is equidistant to both of the sides of the angle.

Converse:

* If it is equidistant to both sides of the angle, then it lies on the angle bisector.

MUST: the point always has to lie on the interior of the angle.


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1.5 cm

1.5 cm

Angle point that Angle bisector

lies on the bisector

is equidistant to both

sides of the angle


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EXAMPLES

Given that BD bisects ∠ABC and

CD = 21, find AD.

AD= 21

2. Given that AD =65 , CD= 65 and m∠ABC = 50˚, find m∠CBD.

m∠DBC= 25˚

3. Given that DA=DC , m∠DBC= (10y+3)˚ , and m ∠DBA= (8y+10) ˚, find m∠DBC.

10y+3=8y+10

2y+3=10

2y=7y=3.5 plug in both equations.

m∠DBC=38˚

A

D

B

C


Journal chapter 5

Option

#3


Journal chapter 5

example. G is the point of concurrency.


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What is a circumcenter?

It is the point where all the 3 perpendicular bisectors of a triangle meet.

What makes it special?

It is the point of concurrency and it is equidistant to all 3 vertices.

1. Knowing what circumcenter is, it would help you if you are a business person to know a strategic place in between 3 cities to place your business. That way it would be located in the same distance from all 3 cities for customers to come.

EXAMPLE

F is the circumcenter in this equilateral triangle. The circumcenter is on the inside of the triangle.


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Examples:

2.In an acute the circumcenter is on the inside of the triangle

C is the circumcenter

3.In a right the circumcenter is on the midpoint of the hypotenuse

D is the circumcenter

4.In an obtuse the circumcenter is on the outside of the triangle.

F is the circumcenter

f

c

d


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Concurrency of Angle bisectors

Of a triangle theorem

 The angle bisectors of a triangle meet at a point that is equidistant from the three sides of the triangle.

What is the distance from H to G, if F to H is 10 cm?

10cm from H to G

2. H to E is 15cm , then what would be H to G doubled?

Real: H to G would be 15cm and doubled would be 30cm

H in this triangle is what?

The incenter of triangle ABC


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Incenter of a

TRIANGLE

  • The point of concurrency where the three angle bisectors of a triangle meet. It is equidistant (same distance) from each three sides of the triangle

EXAMPLES:

2. If you want to open a new restaurant in town you have to place it at the incenter of three major highways so people can go to your restaurant.

3. If the distance from K to P is 2cm, what is the distance from P to L?

From P to L = 2cm

G.

M

K

P

J

L

H

1. P is the incenter


Journal chapter 5

Option

#4


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Median of a triangle

Ex.1 :

Median

Is the segment that goes from the vertex of a triangle to the opposite midpoint

Ex. 2: all medians are congruent. Shown in yellow

Ex.3 :

Median


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Centroid

The point where the medians of a triangle intersect.

Is the center of balance. The distance from the vertex to the centroid is the double the distance from the centroid to the opposite side.

EXAMPLES:

AP= 2/3 AY

BP=2/3BZ

CP=2/3CX

1. Balance a triangular piece of glass on a triangular coffee table.

3.

2. G is the centroid.

b

y

p

x

c

z

a


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Concurrency of Medians of a Triangle Theorem

The medians of a triangle intersect at a point that is 2/3 of the distance from each vertex to the midpoint of the opposite side


Journal chapter 5

Option #5


Journal chapter 5

A segment that goes from the vertex perpendicular to the line containing the opposite side.

ALTITUDE

TRIANGLES !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

hypotenuse

altitude

ORTHOCENTER

Where the altitudes of a triangle intersect.


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Concurrency of Altitudes

of triangles

Concurrency of Altitudes of a Triangle Theorem 

The lines containing the altitudes of a triangle are concurrent


Journal chapter 5

Option

#6


Journal chapter 5

Midsegments:

A segment that joins the midpoints of two sides of the triangle.

Every triangle has 3midsegments that form the midsegment triangle.

Midsegment Theorem  The segment joining the midpoints of two sides of a triangle is parallel to the third side, and is half the length of the third side. 


Journal chapter 5

Option #7


Journal chapter 5

Side – Angle Relationship

In any triangle the longest side is always opposite from the largest angle. The small side is opposite from a small angle.

EXAMPLES:


Journal chapter 5

Option #8


Journal chapter 5

Exterior angle inequality

The exterior angle is bigger (grater) than the non-adjacent interior angles of the triangle .


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TELL WHEATHER A TRIANGLE CAN HAVE THE SIDES WITH THE GIVEN LENGTHS. EXPLAIN

3,5,7 2. 4, 6.5,11 3. 7, 4, 5

3 + 5 > 7 4 +6.5> 11 5+4 > 7

8>7 10.5>119>7

YES, the sum of each NO. by the 3rd triangleYes , 5+4 is grater than 7 so

Pair of the lengths is inequality thm. A trianglethere can be a triangle.

Grater than the 3rd <. Cannot have these side lengths.

TRIANGLE INEQUALITY

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


Journal chapter 5

Option #9


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Howtowriteanindirectproof


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Examples:

PROVE: IF A>0 THAN 1/A >0

Assume a>0 than 1/a<0

(a) 1/a <0a

a 1 <0

CONTRADICTION

3. Therefore if a>0 than 1/a>0

PROVE: A SCALENE TRIANGLE CANNOT HAVE TWO CONGRUENT ANGLES

Assume a scalene triangle can have two congruent sides

a triangle with 2 congruent angles… deff. of an isosceles triangle

Deff. Of isosceles and scalene triangles… therefore a scalene triangle cannot have two congruent angles.

PROVE: A TRIANGLE CANNOT HAVE 2 RIGHT ANGLES

assume a triangle has 2 right angles <1+<2….GIVEN

M<1 =M<2 = 90.. DEFF OF RT. <

M<1+M<2+M<3=180 …..TRIANGLE + THM

M<3= 0 CONTRADICTION

So a triangle cannot have 2 right angles.


Journal chapter 5

Option #19


Journal chapter 5

HINGE THEOREM

If 2 triangles have 2 sides that are congruent, but the third side is not congruent, then the triangle with the large included angle has the longer side.

(What changes is the incidence and the side….) Theorem:

If two sides of one triangle are congruent to the two sides of another triangle and the third sides are not congruent, then the larger included angle is across from the longer third side. Converse :


Journal chapter 5

Option #11


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45˚- 45 ˚- 90˚

90˚- 30˚- 60˚

45

60

Both legs of the triangle are congruent and the length of the hypotenuse is the length of the two legs √2

The length of the hypotenuse is 2x the length of the shorter leg, and the length of the longer leg is the length of the shorter leg time √3

90

45

Examples:

30

90

18√3 =2x

9 √3=x

y=x √3

y=27

60

18√3

x

y

y= 2x

Y=2(5 √3)

Y= 10 √3

x

2 . 15= √3

15/ √3=x

5 √3=x

60

y

30

5

X=5 √3

Y=2(5)

Y=10

30

y


Journal chapter 5

"The only angle from which to approach a problem is the TRY-Angle"


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