Chapter 5 Section 5.4 Medians and Altitudes C C F G M E A H B A B USING MEDIANS OF A TRIANGLE A median of a triangle is a segment whose endpoints are a vertex.

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Transcript Chapter 5 Section 5.4 Medians and Altitudes C C F G M E A H B A B USING MEDIANS OF A TRIANGLE A median of a triangle is a segment whose endpoints are a vertex.

Slide 1

Chapter 5
Section 5.4
Medians and Altitudes

C
C

F

G
M

E

A

H

B

A

B


Slide 2

USING MEDIANS OF A TRIANGLE

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

D


Slide 3

USING MEDIANS OF A TRIANGLE

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

centroid

acute triangle

centroid

centroid

right triangle

obtuse triangle

The medians of a triangle have a special concurrency property.


Slide 4

USING MEDIANS OF A TRIANGLE
THEOREM
THEOREM 5.8 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.
If P is the centroid of ABC, then
2
AD
3
BP = 2 BF
3
CP = 2 CE
3
AP =

P


Slide 5

USING MEDIANS OF A TRIANGLE

The centroid of a triangle
can be used as its
balancing point.

A triangular model of uniform
thickness and density will
balance at the centroid of the triangle.


Slide 6

Using the Centroid of a Triangle

P is the centroid of QRS shown below and PT = 5.
Find RT and RP.
SOLUTION

Because P is the centroid,
RP = 2 RT.
3
1
Then PT = RT – RP = RT
3
1
Substituting 5 for PT, 5 =
RT, so RT = 15.
3
Then RP =

2
RT = 2 (15) = 10.
3
3

So, RP = 10 and RT = 15.


Slide 7

Finding the Centroid of a Triangle

Find the coordinates of the centroid of  JKL.

(5, 8)

SOLUTION

The centroid is two thirds of the
distance from each vertex to the
midpoint of the opposite side.
Choose the median KN. Find
the coordinates of N, the midpoint
of JL.

The coordinates of N are
3 + 7 , 6 + 10
2
2

L

= (5, 8)

(7, 10)

N
P

(5, 2)
K

=

10
, 16
2
2

(3, 6)

J


Slide 8

Finding the Centroid of a Triangle

Find the coordinates of the centroid of  JKL.

(5, 8)

SOLUTION
Find the distance from vertex K

to midpoint N. The distance from
K (5,2) to N(5,8) is 8 – 2, or 6 units.
Determine the coordinates of
2
the centroid, which is • 6, or
3
4 units up from vertex K along

(3, 6)

(7, 10)

N

(5, 6)

P

L

J

M

(5, 2)
K

the median KN.

The coordinates of the centroid P are (5, 2 + 4), or (5, 6)


Slide 9

USING ALTITUDES 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. An altitude can lie inside, on, or outside
the triangle.

Every triangle has three altitudes. The lines containing the
altitudes are concurrent and intersect at a point called
the orthocenter of the triangle.


Slide 10

Drawing Altitudes and Orthocenters

Where is the orthocenter of an acute triangle?
SOLUTION

Draw an example.
The three altitudes intersect at G,
a point inside the triangle.


Slide 11

Drawing Altitudes and Orthocenters

Where is the orthocenter of a right triangle?
SOLUTION

The two legs, LM and KM,
are also altitudes. They intersect
at the triangle’s right angle.

This implies that the orthocenter is on the triangle
at M, the vertex of the right angle of the triangle.


Slide 12

Drawing Altitudes and Orthocenters

Where is the orthocenter of an obtuse triangle?
SOLUTION

The three lines that contain the
altitudes intersect at W,
a point outside the triangle.


Slide 13

USING ALTITUDES OF A TRIANGLE
THEOREM
THEOREM 5.9 Concurrency of Altitudes of a Triangle

The lines containing the altitudes of a triangle
are concurrent.

If AE, BF, and CD are the altitudes
of ABC, then the lines AE, BF, and
CD intersect at some point H.


Slide 14

Median
Angle Bisector

Perpendicular Bisector
Altitude

All Four


Slide 15

2
ML    MP
3
2
10    M P
3
15  M P

PO = 11


Slide 16

2
NL    NQ
3
2
8    NQ
3

12  N Q

N Q  N L  LQ

12  8  LQ
4  LQ