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Lesson 5-1

Lesson 5-1. Bisectors in Triangles. Transparency 5-1. 5-Minute Check on Chapter 4. Refer to the figure. 1. Classify the triangle as scalene, isosceles, or equilateral. 2. Find x if m  A = 10 x + 15, m  B = 8 x – 18, and m  C = 12 x + 3.

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Lesson 5-1

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  1. Lesson 5-1 Bisectors in Triangles

  2. Transparency 5-1 5-Minute Check on Chapter 4 Refer to the figure. 1. Classify the triangle as scalene, isosceles, or equilateral. 2. Find x if mA = 10x + 15, mB = 8x – 18, and mC = 12x + 3. 3. Name the corresponding congruent angles if RSTUVW. 4. Name the corresponding congruent sides if LMNOPQ. 5. Find y if DEF is an equilateral triangle and mF = 8y + 4. 6. What is the slope of a line that contains (–2, 5) and (1, 3)? Standardized Test Practice: 2/3 –3/2 A B C D 3/2 –2/3

  3. Transparency 5-1 5-Minute Check on Chapter 4 Refer to the figure. 1. Classify the triangle as scalene, isosceles, or equilateral. isosceles 2. Find x if mA = 10x + 15, mB = 8x – 18, and mC = 12x + 3. 6 3. Name the corresponding congruent angles if RSTUVW.RU; SV; TW 4. Name the corresponding congruent sides if LMNOPQ. LMOP; MNPQ; LNOQ 5. Find y if DEF is an equilateral triangle and mF = 8y + 4. 7 6. What is the slope of a line that contains (–2, 5) and (1, 3)? Standardized Test Practice: 2/3 –3/2 A B C D 3/2 –2/3

  4. Objectives • Identify and use perpendicular bisectors in triangles • Identify and use angle bisectors in triangles

  5. Vocabulary • Concurrent lines – three or more lines that intersect at a common point • Point of concurrency – the intersection point of three or more lines • Perpendicular bisector – passes through the midpoint of the segment (triangle side) and is perpendicular to the segment • Circumcenter – the point of concurrency of the perpendicular bisectors of a triangle; the center of the largest circle that contains the triangle’s vertices • Incenter – the point of concurrency for the angle bisectors of a triangle; center of the largest circle that can be drawn inside the triangle

  6. Theorems • 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 endpoints of the segments lies on the perpendicular bisector of a segment. • Theorem 5.3, Circumcenter Theorem – The circumcenter of a triangle is equidistant from the vertices of the triangle. • Theorem 5.4 – Any point on the angle bisector is equidistant from the sides of the triangle. • Theorem 5.5 – Any point equidistant from the sides of an angle lies on the angle bisector. • Theorem 5.6, Incenter Theorem – The incenter of a triangle is equidistant from each side of the triangle.

  7. Triangles – Perpendicular Bisectors A Note: from Circumcenter Theorem: AP = BP = CP Midpoint of AC Z Circumcenter P Midpoint of AB X C Midpoint of BC Y B Circumcenter is equidistant from the vertices

  8. Perpendicular Bisector Theorems

  9. Example 1A A. Find BC. From the information in the diagram, we know that line CD is the perpendicular bisector of line segment AB. BC = AC Perpendicular Bisector Theorem BC = 8.5 Substitution Answer:8.5

  10. Example 1B B. Find XY. Answer:6

  11. Example 1C C. Find PQ. From the information in the diagram, we know that line QS is the perpendicular bisector of line segment PR. PQ = RQ Perpendicular Bisector Theorem 3x + 1 = 5x – 3 Substitution 1 = 2x – 3 Subtract 3x from each side. 4 = 2x Add 3 to each side. 2 = x Divide each side by 2. So, PQ = 3(2) + 1 = 7. Answer:7

  12. Perpendicular Bisector Theorems

  13. Example 2 • GARDENA triangular-shaped garden is shown. Can a fountain be placed at the circumcenter and still be inside the garden? By the Circumcenter Theorem, a point equidistant from three points is found by using the perpendicular bisectors of the triangle formed by those points. Copy ΔXYZ, and use a ruler and protractor to draw the perpendicular bisectors. The location for the fountain is C, the circumcenter of ΔXYZ, which lies in the exterior of the triangle. Answer:No, the circumcenter of an obtuse triangle is in the exterior of the triangle.

  14. Triangles – Angle Bisectors A Note: from Incenter Theorem: QX = QY = QZ Z Q Incenter C X Y B Incenter is equidistant from the sides

  15. Angle Bisector Theorems

  16. Example 3A A. Find DB. From the information in the diagram, we know that ray AD is the angle bisector of  BAC. DB = DC Angle Bisector Theorem DB = 5 Substitution Answer:5

  17. Example 3B B. Find mWYZ. Answer:mWYZ = 28°

  18. Example 3C C. Find QS. QS = SR Angle Bisector Theorem 4x – 1 = 3x + 2 Substitution x – 1 = 2 Subtract 3x from each side. x = 3 Add 1 to each side. Answer:QS = 4(3) – 1 = 11

  19. Angle Bisector Theorems

  20. Example 4A A. Find ST if S is the incenter of ΔMNP. By the Incenter Theorem, since S is equidistant from the sides of ΔMNP,ST = SU. Find ST by using the Pythagorean Theorem. a2 + b2 = c2 Pythagorean Theorem 82 + SU2 = 102 Substitution 64 + SU2 = 100 82 = 64, 102 = 100 SU2 = 36 subtract 36 from both sides SU = 6 positive square root Answer:ST = SU = 6

  21. Example 4B B. Find mSPU if S is the incenter of ΔMNP. Since NR and MU are angle bisectors, we double the given half-angles to get the whole angles in: mUPR + mRMT + mTNU = 180 Triangle Angle Sum Theorem mUPR + 62 + 56 = 180 Substitution mUPR + 118 = 180 Simplify. mUPR = 62 Subtract 118 from each side. Answer: mSPU = (1/2) mUPR = 31

  22. Given: Example 5 mDGE Find: 2x + 80 + 30 = 180 3 ’s in triangle sum to 180 2x + 110 = 180 2x = 70 subtracting 110 from both sides x = 35 dividing both sides by 2 x + mDGE + 30 = 180 3 ’s in triangle sum to 180 35 + mDGE + 30 = 180 substitute x mDGE + 65 = 180 combine numbers mDGE = 115 subtract 65 from both sides

  23. Special Segments in Triangles

  24. Location of Point of Concurrency

  25. Summary & Homework • Summary: • Perpendicular bisectors and angle bisectors of a triangle are all special segments in triangles • Perpendiculars bisectors: • form right angles • divide a segment in half – go through midpoints • equal distance from the vertexes of the triangle • Angle bisector: • cuts angle in half • equal distance from the sides of the triangle • Homework: • pg 327-31; 1-3, 5-7, 9-11, 26-30

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