Transcript Lesson
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 mA = 10x + 15, mB = 8x – 18, and mC = 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 mF = 8y + 4. 6. Standardized Test Practice: What is the slope of a line that contains (–2, 5) and (1, 3)? A –2/3 B 2/3 C –3/2 D 3/2 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 mA = 10x + 15, mB = 8x – 18, and mC = 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 mF = 8y + 4. 6. Standardized Test Practice: What is the slope of a line that contains (–2, 5) and (1, 3)? A –2/3 B 2/3 7 C –3/2 D 3/2 Objectives • Identify and use perpendicular bisectors in triangles • Identify and use angle bisectors in triangles 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 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. Triangles – Perpendicular Bisectors A Note: from Circumcenter Theorem: AP = BP = CP Z Midpoint of AC Circumcenter Midpoint of AB X P C Y Midpoint of BC B Circumcenter is equidistant from the vertices Perpendicular Bisector Theorems 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 Example 1B B. Find XY. Answer: 6 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 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 Perpendicular Bisector Theorem Perpendicular Bisector Theorems Example 2 • GARDEN A 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. Triangles – Angle Bisectors A Note: from Incenter Theorem: QX = QY = QZ Z Q Incenter C X Y B Incenter is equidistant from the sides Angle Bisector Theorems 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 Example 3B B. Find mWYZ. Answer: mWYZ = 28° Example 3C C. Find QS. QS = SR 4x – 1 = 3x + 2 x–1 =2 x =3 Angle Bisector Theorem Substitution Subtract 3x from each side. Add 1 to each side. Answer: QS = 4(3) – 1 = 11 Angle Bisector Theorems 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 SU = 6 Answer: ST = SU = 6 subtract 36 from both sides positive square root Example 4B B. Find mSPU 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: mUPR + mRMT + mTNU = 180 mUPR + 62 + 56 = 180 mUPR + 118 = 180 mUPR = 62 Triangle Angle Sum Theorem Substitution Simplify. Subtract 118 from each side. Answer: mSPU = (1/2) mUPR = 31 Example 5 Given: Find: mDGE 2x + 80 + 30 = 180 2x + 110 = 180 2x = 70 x = 35 x + mDGE + 30 = 180 35 + mDGE + 30 = 180 mDGE + 65 = 180 mDGE = 115 3 ’s in triangle sum to 180 subtracting 110 from both sides dividing both sides by 2 3 ’s in triangle sum to 180 substitute x combine numbers subtract 65 from both sides Special Segments in Triangles Name Type Point of Concurrency Perpendicular Line, Circumcenter bisector segment or ray Angle bisector Line, segment or ray Incenter Center Special Quality From / To Equidistant from vertices None midpoint of segment Equidistant from sides Vertex none Location of Point of Concurrency Name Point of Concurrency Perpendicular bisector Circumcenter Angle bisector Incenter Triangle Classification Acute Right Obtuse Inside hypotenuse Outside Inside Inside Inside 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