1. 5.4 Use Medians and Altitudes

2. Median of a triangle
Median: segment whose endpoints are the vertex of the triangle and the midpoint of the opposite side.

3. The centroid is the balancing point of a triangle.
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

4. 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 X W D C B Y

5. Use the centroid of a triangle
EXAMPLE 1
Use the centroid of a triangle
In RST, Q is the centroid and SQ = 8. Find QW and SW.
SOLUTION
SQ = 2 3 SW
Concurrency of Medians of a Triangle Theorem
8 = 2 3 SW
Substitute 8 for SQ.
Multiply each side by the reciprocal, . 2 3
12 = SW
Then QW =
SW – SQ =
12 – 8 = 4.
So, QW = 4 and SW = 12.

6. Median of a triangle
Median: segment whose endpoints are the vertex of the triangle and the midpoint of the opposite side. ? 10
Centroid 3 6 5

7. Incenter Centroid Circumcenter Outside if obtuse. On hyp.if right 
 bisectors
Equidistant from each vertex.
Circumscribe circle outside the triangle.
Incenter
(always inside the )
Angle bisectors
Equidistant from each side of the .
Inscribe circle within the triangle.
Centroid
Medians (from vertex to opposite midpoint)
Vertex to centroid is 2/3 of length.
Balancing point of the triangle.

8. L is the centroid of MNO NP = 11, ML = 10, NL = 8
PO = _____

9. L is the centroid of MNO NP = 11, ML = 10, NL = 815
MP = _____ 5

10. L is the centroid of MNO NP = 11, ML = 10, NL = 84
LQ = _____

11. L is the centroid of MNO NP = 11, ML = 10, NL = 8
Perimeter of NLP = _____ 24

12. Find the coordinates of D, the midpoint of segment AB.
(6,3)

13. Find the length of median CD.
(6,3)

14. 8 3

15. Find VZ 4 8 3

16. 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.

17. Concurrency of Altitudes of a 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.

18. EXAMPLE 3
Find the orthocenter
Find the orthocenter P in an acute, a right, and an obtuse triangle.
SOLUTION
Acute triangle
P is inside triangle.
Right triangle
P is on triangle.
Obtuse triangle
P is outside triangle.

19. 3 2 3 and 4 4 1, 2, 3, and 4 1 =  bisector 2 = angle bisector
3 = a median
4 = an altitude
3 and 4 2 4
1, 2, 3, and 4

20. 4
1 =  bisector
2 = angle bisector
3 = a median
4 = an altitude

21. 2
1 =  bisector
2 = angle bisector
3 = a median
4 = an altitude

22. 3
1 =  bisector
2 = angle bisector
3 = a median
4 = an altitude

23. 1
1 =  bisector
2 = angle bisector
3 = a median
4 = an altitude

24. Assignment #41: Page 322 # 3 – 27, 33 – 35