5.4 Use Medians and
Altitudes
Median
of a triangle
Median: segment
whose endpoints are the
vertex of the triangle
and the midpoint of the
opposite side.
Median: segment whose endpoints are the
vertex of the triangle and the midpoint of
the opposite side.
The centroid is the
balancing point of a
triangle.
Centroid
Theorem 5.8
Concurrency of Medians
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.
A
WD 
1
2
W
B
X
D
Y
 DC 
XD 
1
2
 BD 
C
EXAMPLE 1
Use the centroid of a triangle
In RST, Q is the centroid and SQ = 8. Find QW and
SW.
SOLUTION
SQ =
2
SW
3
8 = 2 SW
3
12 = SW
Concurrency of Medians of a
Triangle Theorem
Substitute 8 for SQ.
Multiply each side by the reciprocal, 2 .
3
Then QW = SW – SQ = 12 – 8 =4.
So, QW = 4 and SW = 12.
Median: segment
whose endpoints are the
vertex of the triangle
and the midpoint of the
opposite side.
Median
of a triangle
Centroid
?
10
3
6
5
Circumcenter  bisectors
Outside if
obtuse. On
hyp.if right 
Equidistant
from each
vertex.
Incenter
Equidistant
Inscribe
from each
circle within
side of the . the triangle.
Angle
(always inside bisectors
the )
Circumscribe
circle outside
the triangle.
Vertex to
Balancing
Medians
point of the
(always inside (from vertex centroid is
to opposite 2/3 of length. triangle.
the )
midpoint)
Centroid
L is the centroid of MNO
NP = 11, ML = 10, NL = 8
11
PO = _____
L is the centroid of MNO
NP = 11, ML = 10, NL = 8
15
MP = _____
5
L is the centroid of MNO
NP = 11, ML = 10, NL = 8
4
LQ = _____
L is the centroid of MNO
NP = 11, ML = 10, NL = 8
Perimeter of
24
NLP = _____
Find the coordinates of D, the
midpoint of segment AB.
x
48
6
2
y 
51
2
(6,3)
3
Find the length of median CD.
CD 
(6,3)
2
4 6
2
CD 
CD 
52
4  13
C D  2 13
2
8
3
Find VZ
4
8
3
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.
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.
EXAMPLE 3
Find the orthocenter
Find the orthocenter P in an acute, a right,
and an obtuse triangle.
SOLUTION
Acute triangle
Right triangle
Obtuse triangle
P is inside triangle. P is on triangle. P is outside triangle.
1 =  bisector
3
2
3 and 4
2 = angle bisector
3 = a median
4
1, 2, 3, and 4
4 = an altitude
 BAE   EAC and BF  FC
4
AD is a ___
1 =  bisector
2 = angle bisector
3 = a median
4 = an altitude
 BAE   EAC and BF  FC
2
AE is a ___
1 =  bisector
2 = angle bisector
3 = a median
4 = an altitude
 BAE   EAC and BF  FC
3
AF is a ___
1 =  bisector
2 = angle bisector
3 = a median
4 = an altitude
 BAE   EAC and BF  FC
1
GF is a ___
1 =  bisector
2 = angle bisector
3 = a median
4 = an altitude
Assignment #41:
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