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5.1 Bisectors, Medians, and Altitudes

5.1 Bisectors, Medians, and Altitudes. Objectives. Identify and use ┴ bisectors and  bisectors in ∆s Identify and use medians and altitudes in ∆s. C. Side AB. A. P. B. perpendicular bisector. Perpendicular Bisector. A ┴ bisector of a ∆ is a line, segment, or

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5.1 Bisectors, Medians, and Altitudes

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  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. C Side AB A P B perpendicular bisector Perpendicular Bisector A ┴ bisectorof a ∆ is a line, segment, or ray that passes through the midpoint of one of the sides of the ∆ at a 90° .

  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… C Side AB A P B CP is perpendicular bisector 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 C D

  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. Example 1: 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.

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

  19. Example 1: 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:

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

  21. Example 2: COORDINATE GEOMETRY The vertices of HIJare H(1, 2), I(–3, –3), andJ(–5, 1). Find the coordinates of the orthocenter of HIJ.

  22. Find an equation of the altitude from The slope of so the slope of an altitude is Example 2: Point-slope form Distributive Property Add 1 to each side.

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

  24. Substitution, Example 2: Then, solve a system of equations to find the point of intersection of the altitudes. 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.

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

  26. Your Turn: COORDINATE GEOMETRY The vertices of ABCare A(–2, 2), B(4, 4), and C(1, –2). Find the coordinates of the orthocenter of ABC. Answer: (0, 1)

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

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