# Bisectors, Medians and Altitudes - PowerPoint PPT Presentation

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## Bisectors, Medians and Altitudes

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### Lesson 5-1 Bisectors, Medians and Altitudes Objectives Identify and use perpendicular bisectors and angle bisectors in triangles Identify and use medians and ... – PowerPoint PPT presentation

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Title: Bisectors, Medians and Altitudes

1
Lesson 5-1
• Bisectors, Medians and Altitudes

2
Objectives
• Identify and use perpendicular bisectors and
angle bisectors in triangles
• Identify and use medians and altitudes in
triangles

3
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
• Median segment whose endpoints are a vertex of
a triangle and the midpoint of the side opposite
the vertex
• Altitude a segment from a vertex to the line
containing the opposite side and perpendicular to
the line containing that side

4
Vocabulary
• Circumcenter the point of concurrency of the
perpendicular bisectors of a triangle the center
of the largest circle that contains the
triangles vertices
• Centroid the point of concurrency for the
medians of a triangle point of balance for any
triangle
• Incenter the point of concurrency for the angle
bisectors of a triangle center of the largest
circle that can be drawn inside the triangle
• Orthocenter intersection point of the
altitudes of a triangle no special significance

5
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.
• Theorem 5.7, Centroid Theorem The centroid of
a triangle is located two thirds of the distance
from a vertex to the midpoint of the side
opposite the vertex on a median.

6
Triangles Perpendicular Bisectors
A
Note from Circumcenter Theorem AP BP CP
Midpoint of AC
Z
Circumcenter
P
Midpoint of AB
X
C
Midpoint of BC
Y
B
Circumcenter is equidistant from the vertices
7
Triangles Angle Bisectors
A
Note from Incenter Theorem QX QY QZ
Z
Q
Incenter
C
X
Y
B
Incenter is equidistant from the sides
8
Triangles Medians
A
Note from Centroid theorem BM 2/3 BZ
Midpoint of AC
Z
Midpoint of AB
Centroid
X
M
C
Medianfrom B
Y
Midpoint of BC
B
Centroid is the point of balance in any triangle
9
Triangles Altitudes
A
Note Altitude is the shortest distance from a
vertex to the line opposite it
Z
Altitudefrom B
C
Orthocenter
X
Y
B
Orthocenter has no special significance for us
10
Special Segments in Triangles

Name Type Point of Concurrency Center SpecialQuality From / To
Perpendicular bisector Line, segment or ray Circumcenter Equidistantfrom vertices Nonemidpoint of segment
Angle bisector Line, segment or ray Incenter Equidistantfrom sides Vertexnone
Median segment Centroid Center ofGravity Vertexmidpoint of segment
Altitude segment Orthocenter none Vertexnone
11
Location of Point of Concurrency
Name Point of Concurrency Triangle Classification Triangle Classification Triangle Classification
Name Point of Concurrency Acute Right Obtuse
Perpendicular bisector Circumcenter Inside hypotenuse Outside
Angle bisector Incenter Inside Inside Inside
Median Centroid Inside Inside Inside
Altitude Orthocenter Inside Vertex - 90 Outside
12
Find
m?DGE
13
EXAMPLE 2
Find
14
EXAMPLE 3
ALGEBRA Points U, V, and W are the midpoints of
respectively. Find a, b, and c.
Find a.
Centroid Theorem
Substitution
Multiply each side by 3 and simplify.
Subtract 14.8 from each side.
Divide each side by 4.
15
CONT.
Find b.
Centroid Theorem
Substitution
Multiply each side by 3 and simplify.
Subtract 6b from each side.
Subtract 6 from each side.
Divide each side by 3.
16
CONT.
Find c.
Centroid Theorem
Substitution
Multiply each side by 3 and simplify.
Subtract 30.4 from each side.
Divide each side by 10.