1.  Perpendicular Bisector- a line, segment, or ray that passes through the midpoint of the side and is perpendicular to that side  Theorem 5.1  Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.  Theorem 5.2  Any point equidistant from the endpoint s of a segment lies on the perpendicular bisector of the segment

2.  Concurrent lines- when 3 or more lines intersect at a common point  Point of Concurrency- the point that the concurrent lines intersect at  Circumcenter- the point of concurrency of the perpendicular bisectors of a triangle  Theorem 5-3 Circumcenter Theorem  The circumcenter of a triangle is equidistant from the vertices of a triangle

3.  Theorem 5.4  Any point on the angle bisector is equidistant from the sides of the angle  Theorem 5.5  Any point equidistant from the sides of an angle lies on the angle bisector  Incenter- the point of concurrency of the angle bisectors of a triangle.  Theorem 5.6 Incenter Theorem  The incenter of a triangle is equidistant from each side of the triangle

4.  Median- segment whose endpoints are a vertex of a triangle and the midpoint of the side opposite the vertex  Centroid- the point of concurrency for the medians of a triangle.  Theorem 5.7- Centroid Theorem  The centroid of a triangle is located two thirds of the distance from a vertex to the midpoint of the side opposite the vertex on the median

5.  Altitude- segment from the vertex to the line containing the opposite side and perpendicular to the line containing that side  Orthocenter- the intersection point of the altitudes of a triangle

6. Given: Prove:

7. Proof: Statements Reasons 1. 1. Given 2. 2. Angle Sum Theorem 3. 3. Substitution 4. 4. Subtraction Property 5. 5. Definition of angle bisector 6. 6. Angle Sum Theorem 7. 7. Substitution 8. 8. Subtraction Property

8. Prove: Given:.

9. Proof: Statements. Reasons 1. Given 2. Angle Sum Theorem 3. Substitution 4. Subtraction Property 5. Definition of angle bisector 6. Angle Sum Theorem 7. Substitution 8. Subtraction Property 1. 2. 3. 4. 5. 6. 7. 8. 1. Given

10. ALGEBRA Points U, V, and W are the midpoints of respectively. Find a, b, and c.

11. 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.

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

13. 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:

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

15. 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.

16. 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.

17. 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.

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

19. 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:

20. 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)