Example 1C: Applying the Perpendicular Bisector Theorem and Its Converse

2. Example 1C: Applying the Perpendicular Bisector Theorem and Its Converse Find each measure. TU TU = UV  Bisector Thm. 3x + 9 = 7x – 17 Substitute the given values. 9 = 4x – 17 Subtract 3x from both sides. 26 = 4x Add 17 to both sides. 6.5 = x Divide both sides by 4. So TU = 3(6.5) + 9 = 28.5.

4. Since, JM = LM, and , bisects JKL by the Converse of the Angle Bisector Theorem. Example 2C: Applying the Angle Bisector Theorem Find mMKL. mMKL = mJKM Def. of  bisector 3a + 20 = 2a + 26 Substitute the given values. a + 20 = 26 Subtract 2a from both sides. a= 6 Subtract 20 from both sides. So mMKL = [2(6) + 26]° = 38°

5. A median of a triangleis a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side. Every triangle has three medians, and the medians are concurrent.

6. The point of concurrency of the medians of a triangle is the centroid of the triangle. The centroid is always inside the triangle. The centroid is also called the center of gravity because it is the point where a triangular region will balance.

7. Example 1A: Using the Centroid to Find Segment Lengths In ∆LMN, RL = 21 and SQ =4. Find LS. Centroid Thm. Substitute 21 for RL. LS = 14 Simplify.

8. An altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side. Every triangle has three altitudes. An altitude can be inside, outside, or on the triangle.

9. Example : Finding the altitudes Find the altitudes of ∆XYZ with vertices X(0,0), Y(4,4), and Z(6,0). Step 1 Graph the triangle.

10. Step 2 Find an equation of the line containing the altitude from Y to XZ. Since XZ is horizontal, the altitude is vertical. The line containing it must pass through Y(4, 4) so the equation of the line is X = 4. Example Continued

11. Step 3 Find an equation of the line containing the altitude from X to YZ. The slope of a line perpendicular to YZ is . This line must pass through Y(0, 0). Example Continued

12. Step 4 Find an equation of the line containing the altitude from Z to XY. The slope of a line perpendicular to XY is -1 . This line must pass through Y(6, 0). Example Continued

13. A midsegment of a triangleis a segment that joins the midpoints of two sides of the triangle. Every triangle has three midsegments, which form the midsegment triangle.

14. The vertices of ΔRST are R(–7, 0), S(–3, 6), and T(9, 2). M is the midpoint of RT, and N is the midpoint of ST. Show that and Check It Out! Example 1 Step 1 Find the coordinates of M and N.

15. Step 2 Compare the slopes of MN and RS. Since the slopes are equal . Check It Out! Example 1 Continued

16. Step 3 Compare the heights of MN and RS. The length of MN is half the length of RS. Check It Out! Example 1 Continued

17. The relationship shown in Example 1 is true for the three midsegments of every triangle.

18. Use the diagram for Items 3–4. 3. Given that FH is the perpendicular bisector of EG, EF = 4y – 3, and FG = 6y – 37, find FG. 4. Given that EF = 10.6, EH = 4.3, and FG = 10.6, find EG Lesson Quiz: Part I Use the diagram for Items 1–2. 1. Given that mABD = 16°, find mABC. 2. Given that mABD = (2x + 12)° and mCBD = (6x – 18)°, find mABC. 32° 54° 65 8.6

19. Lesson Quiz: Part II 5. Find the value of n. 6.∆XYZ is the midsegment triangle of∆WUV. What is the perimeter of∆XYZ? 16 11.5