5.4  Use Medians and Altitudes

1. 5.4 Use Medians and Altitudes

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

3. Centroid 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.

4. Theorem 5.8Concurrency of MediansThe 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. In RST, Qis the centroid and SQ = 8. Find QWand SW. SQ = SW 2 2 3 3 8= SW 2 3 Multiply each side by the reciprocal, . 12= SW 12 –8 = 4. SW – SQ = Then QW = EXAMPLE 1 Use the centroid of a triangle SOLUTION Concurrency of Medians of a Triangle Theorem Substitute 8 for SQ. So, QW = 4 and SW = 12.

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

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

9. L is the centroid of MNONP = 11, ML = 10, NL = 8 15 MP = _____ 5

10. L is the centroid of MNONP = 11, ML = 10, NL = 8 4 LQ = _____

11. L is the centroid of MNONP = 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. 3 8

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 orthocenterof the triangle.

17. THEOREM 5.9 If AE, BF, and CD are the altitudes of ABC, then the lines AE, BF, and CD intersect at some point H. Concurrency of Altitudes of a Triangle The lines containing the altitudes of a triangle are concurrent.

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

19. 3 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