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Bellwork For the following problems, use A(5,10), B(2,10), C(3,3) Find AB Find the midpoint of CA Find the midpoint of AB Find the slope of AB

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Bellwork. For the following problems, use A(5,10), B(2,10), C(3,3) Find AB Find the midpoint of CA Find the midpoint of AB Find the slope of AB. Bellwork Solution. For the following problems, use A(5,10), B(2,10), C(3,3) Find AB. Bellwork Solution. - PowerPoint PPT Presentation

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Page 1: Bellwork

BellworkFor the following problems, use A(5,10), B(2,10), C(3,3)•Find AB•Find the midpoint of CA•Find the midpoint of AB•Find the slope of AB

Page 2: Bellwork

Bellwork SolutionFor the following problems, use A(5,10), B(2,10), C(3,3)•Find AB

Page 3: Bellwork

Bellwork SolutionFor the following problems, use A(5,10), B(2,10), C(3,3) Find the midpoint of CA

Page 4: Bellwork

Bellwork SolutionFor the following problems, use A(5,10), B(2,10), C(3,3)•Find the midpoint of AB

Page 5: Bellwork

Bellwork SolutionFor the following problems, use A(5,10), B(2,10), C(3,3)•Find the slope of AB

Page 6: Bellwork

Use Perpendicular BisectorsSection 5.2

Use Angle Bisectors of Triangles

Section 5.3

Use Medians and Altitudes

Section 5.4

Page 7: Bellwork

The ConceptYesterday we investigated the concept of circumcentersToday we’re going to look at a couple of other kinds of

points of concurrencyWe’ll look at each of the theorems that helps us to

create these points, while also enhancing our understanding of triangles

Page 8: Bellwork

Isosceles TrianglesWhat is the hallmark of an Isosceles Triangle?

A

Isosceles triangles can also be made by combining two right triangles

A

This creates a situation in which the base is twice the size of the previous triangle, as well the two end vertices are equidistant to the top vertex

Page 9: Bellwork

Perpendicular BisectorsThis isosceles triangle is also created when a perpendicular

bisector extends from a point to a line segment

A A

This gives us the Perpendicular Bisector Theorem

B B

Page 10: Bellwork

TheoremsTheorem 5.2: Perpendicular Bisector Theorem

In a plane, if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.

C

B

Theorem 5.3: Converse of the Perpendicular Bisector TheoremIn a plane, if a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment

A

Page 11: Bellwork

ExampleFind the length of segment AB

5x

CD

B

A

4x+3

5 4 33

x xx

5(3) 15AB

Page 12: Bellwork

TheoremTheorem 5.4: Concurrency of Perpendicular Bisectors of a

TriangleThe perpendicular bisectors of a triangle intersect at a point that is equidistant from the vertices of the triangle

This point is called the circumcenter which means all circles created from it will include all of the vertices

Page 13: Bellwork

TheoremsTheorem 5.5: Angle Bisector Theorem

If a point is on the bisector of an angle, then it is equidistant from the two sides of the angle.

C

B

Theorem 5.6: Converse of the Angle Bisector TheoremIf a point is in the interior of an angle and is equidistant from the sides of an angle, then it lies on the bisector of the angle

AD

Page 14: Bellwork

ExampleFind x

C

B

AD

x

15

27o

27o

Page 15: Bellwork

ExampleFind x & y

35 4 136 49

xxx

C

B

AD

5y-8

3y+12

35o

4x-1o

5 8 3 122 8 122 20

10

y yyyy

Page 16: Bellwork

TheoremTheorem 5.7: Concurrency of Angle Bisectors of a Triangle

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

This point of concurrency is

called the incenter. Circles centered at this point will equally touch all three sides of the

triangle

Page 17: Bellwork

Theorems

Theorem 5.8: Concurrency of Medians of a TriangleThe 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

Median: Segment from the vertex of a triangle to the midpoint of the opposite side

This point of concurrency is

called the centroid. The centroid is the

center of area of the object. It’s the point on which we could balance the

entire object

Page 18: Bellwork

ExampleFind x

10

x

Page 19: Bellwork

ExampleFind x & y if EB=15 and DA=14

A

2x-1

BC

D

E

F

G y+2

3y

15 3(2 1)15 6 318 63

xxxx

14 3 214 4 212 43

y yyyy

Page 20: Bellwork

DefinitionAltitude:

Perpendicular segment from a vertex to the opposite side or to the line that contains the opposite side

Page 21: Bellwork

TheoremsTheorem 5.9: Concurrency of Altitudes of a TriangleThe lines containing the altitudes of a triangle are concurrent

Orthocenter

Page 22: Bellwork

Homework

5.2 Exercises1-10, 16, 17, 245.3 Exercises1, 2-20 even, 285.4 Exercises1-8, 13-15, 37

Page 23: Bellwork

Most Important Points• Perpendicular Bisector Theorem• Circumcenters• Angle Bisector Theorem• Incenters• Medians• Centroids• Altitudes• Orthocenters