Transcript Slide 1

5.1
Bicycles,
Mediums,
and
Attitudes
5.1
Bisectors,
Medians,
and
Altitudes
Objectives

Identify and use ┴ bisectors and  bisectors
in ∆s

Identify and use medians and altitudes in ∆s
Perpendicular Bisector
A ┴ bisector of a ∆ is
a line, segment, or
ray that passes
through the midpoint
of one of the sides of
the ∆ at a 90° .
C
Side AB
A
P
B
perpendicular bisector
┴ Bisector Theorems

Theorem 5.1 – Any point on the ┴ bisector
of a segment is equidistant from the
endpoints of the segment.

Theorem 5.2 – Any point equidistant from
the endpoints of a segment lies on the ┴
bisector of the segment.
┴ Bisector Theorems (continued)
C
Basically, if CP is
the perpendicular
bisector of AB,
then PA ≅ PB.
Side AB
A
P
B
perpendicular bisector
┴ Bisector Theorems (continued)

Since there are three
sides in a ∆, then
there are three ┴
Bisectors in a ∆.

These three ┴
bisectors in a ∆
intersect at a common
point called the
circumcenter.
┴ Bisector Theorems (continued)

Theorem 5.3 (Circumcenter Theorem)
The circumcenter of a ∆ is equidistant from the vertices of
the ∆.
circumcenter

Notice, a circumcenter of a ∆ is the center of the circle we
would draw if we connected all of the vertices with a circle
on the outside (circumscribe the ∆).
Angle Bisectors of ∆s

Another special bisector which we have already
studied is an  bisector. As we have learned, an 
bisector divides an  into two ≅ parts. In a ∆, an 
bisector divides one of the ∆s s into two ≅ s.
(i.e. if AD is an  bisector then BAD ≅ CAD)
B
D
C
Angle Bisectors of ∆s (continued)

Theorem 5.4 – Any point on an  bisector is
equidistant from the sides of the .

Theorem 5.5 – Any point equidistant from the
sides of an  lies on the  bisector.
Angle Bisectors of ∆s (continued)

As with ┴ bisectors, there are three 
bisectors in any ∆. These three  bisectors
intersect at a common point we call the
incenter.
incenter
Angle Bisectors of ∆s (continued)

Theorem 5.6 (Incenter Theorem)
The incenter of a ∆ is equidistant from each
side of the ∆.
Medians

A median is a segment
whose endpoints are a
vertex of a ∆ and the
midpoint of the side
opposite the vertex. Every
∆ has three medians.

These medians intersect
at a common point called
the centroid.

The centroid is the point
of balance for a ∆.
Medians (continued)

Theorem 5.7
(Centroid Theorem)
The centroid of a ∆ is
located two thirds of
the distance from a
vertex to the midpoint
of the side opposite
the vertex on a
median.
Altitudes

An altitude of a ∆ is a segment from a
vertex to the line containing the opposite
side and ┴ to the line containing that side.
Every ∆ has three altitudes.

The intersection point of the altitudes of a ∆
is called the orthocenter.
Altitudes (continued)
Altitude
Orthocenter
Example 1:
ALGEBRA: Points U, V, and W are the midpoints of
respectively. Find a, b, and c.
Example 1:
Find a.
Segment Addition Postulate
Centroid Theorem
Substitution
Multiply each side by 3 and simplify.
Subtract 14.8 from each side.
Divide each side by 4.
Example 1:
Find b.
Segment Addition Postulate
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.
Example 1:
Find c.
Segment Addition Postulate
Centroid Theorem
Substitution
Multiply each side by 3 and simplify.
Subtract 30.4 from each side.
Divide each side by 10.
Answer:
Your Turn:
ALGEBRA: Points T, H, and G are the midpoints of
respectively. Find w, x, and y.
Answer:
Example 2:
COORDINATE GEOMETRY The vertices of HIJ are
H(1, 2), I(–3, –3), and J(–5, 1). Find the coordinates of
the orthocenter of HIJ.
Example 2:
Find an equation of the altitude from
The slope of
so the slope of an altitude is
Point-slope form
Distributive Property
Add 1 to each side.
Example 2:
Next, find an equation of the altitude from I to
slope of
The
so the slope of an altitude is –6.
Point-slope form
Distributive Property
Subtract 3 from each side.
Example 2:
Then, solve a system of equations to find the point
of intersection of the altitudes.
Equation of altitude from J
Substitution,
Multiply each side by 5.
Add 105 to each side.
Add 4x to each side.
Divide each side
by –26.
Example 2:
Replace x with
y-coordinate.
in one of the equations to find the
Rename as improper fractions.
Multiply and simplify.
Answer: The coordinates of the orthocenter of
Your Turn:
COORDINATE GEOMETRY The vertices of ABC are
A(–2, 2), B(4, 4), and C(1, –2). Find the coordinates of
the orthocenter of ABC.
Answer: (0, 1)
Assignment

Geometry:
Pg. 242 #6 – 9, 13 – 30

Pre - AP Geometry:
Pg. 242 #6 – 9, 13 – 20, 27 - 30