1. 5.1 Bisectors, Medians, and Altitudes

2. Objectives Identify and use ┴ bisectors and  bisectors in ∆s Identify and use medians and altitudes in ∆s

3. 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° . Side AB perpendicular bisector PAB C

4. ┴ 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.

5. ┴ Bisector Theorems (continued) Basically, if CP is the perpendicular bisector of AB, then… Side AB CP is perpendicular bisector PAB C CA ≅ CB.

6. ┴ 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.

7. ┴ Bisector Theorems (continued) Theorem 5.3 (Circumcenter Theorem) The circumcenter of a ∆ is equidistant from the vertices of the ∆. 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 ∆). circumcenter

8. 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

9. 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.

10. 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

11. Angle Bisectors of ∆s (continued) Theorem 5.6 (Incenter Theorem) The incenter of a ∆ is equidistant from each side of the ∆.

12. 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 ∆.

13. 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.

14. 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.

15. Altitudes (continued) Altitude Orthocenter

16. ALGEBRA: Points U, V, and W are the midpoints of respectively. Find a, b, and c. Example 1:

17. 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:

18. Find b. Segment Addition Postulate Centroid Theorem Substitution Multiply each side by 3 and simplify. Subtract 6b from each side. Divide each side by 3. Subtract 6 from each side. Example 1:

19. 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: Example 1:

20. ALGEBRA: Points T, H, and G are the midpoints of respectively. Find w, x, and y. Answer: Your Turn:

21. 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:

22. 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:

23. Point-slope form Distributive Property Subtract 3 from each side. Next, find an equation of the altitude from I to The slope of so the slope of an altitude is –6. Example 2:

24. Equation of altitude from J Multiply each side by 5. Add 105 to each side. Add 4x to each side. Divide each side by –26. Substitution, Then, solve a system of equations to find the point of intersection of the altitudes. Example 2:

25. Replace x with in one of the equations to find the y-coordinate. Multiply and simplify. Rename as improper fractions. The coordinates of the orthocenter of Answer: Example 2:

26. 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) Your Turn:

27. Assignment Geometry: Pg. 242 #6, 13 – 26 Pre - AP Geometry: Pg. 242 #6 – 9, 13 – 30