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Points of Concurrency in Triangles (5-1 through 5 – 4)

Points of Concurrency in Triangles (5-1 through 5 – 4). Essential Question – What special segments exist in triangles?. 5-2 Use Perpendicular Bisectors. TBK p.264. Perpendicular bisector – a segment, ray, line, or plane that is perpendicular to a segment at its midpoint

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Points of Concurrency in Triangles (5-1 through 5 – 4)

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  1. Points of Concurrency in Triangles(5-1 through 5 – 4) Essential Question – What special segments exist in triangles?

  2. 5-2 Use Perpendicular Bisectors TBK p.264 • Perpendicular bisector – a segment, ray, line, or plane that is perpendicular to a segment at its midpoint • Equidistant - same distance • Circumcenter – the point of concurrency of the 3 perpendicular bisectors of a triangle

  3. THEOREM 5.2: Perpendicular Bisector Theorem • If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. • If CP is the perp bisector of AB, then CA = CB • THEOREM 5.3: Converse of the Perpendicular Bisector Theorem • If a point is equidistant from the endpoints of the segment, then it is on the perpendicular bisector of the segment. • If DA = DB, then D lies on the CP. perp bisector

  4. TBK p.264 JM = ML  Bisector Thm 7x = 3x + 16 Substitute – 3x = – 3x Solve for x 4x = 16 4 = 4 x = 4 ML = 3x + 16 = 3(4) + 16 = 12 + 16 = 28

  5. TBK p.265 AC = CB DC bisects AB AE = BE Definition of Congruence AD = BD  Bisector Thm. Yes, because AE = BE, E is equidistant from B and A. According to Converse of  Bisector, E is on the  Bisector DC.

  6. KG = KH , JG = JH , FG = FH GH = KG + KH KG = KH GH = 2x + (x + 1) 2x = x + 1 -x -x GH = 2(1)+ (1 + 1) GH = 2+ (2) x = 1 GH = 4

  7. Do the perpendicular bisector part of the task • Then show the geosketch example, showing how the center can move depending on the type of triangle used.

  8. The perpendicular bisectors are concurrent at a point called the circumcenter. Note: The circumcenter can be inside or outside of the triangle Note: The circumcenter is equal distance from all the vertices.

  9. 5.3 Use Angle Bisectors of Triangles NTG p.282 • Angle bisector – a ray that divides an angle into two congruent adjacent angles • Incenter– point of concurrency of the three angle bisectors **NOTE: In geometry, distance means the shortest length between two objects and this is always perpendicular. **

  10. THEOREM 5.5: ANGLE BISECTOR THEOREM If a point is on the bisector of an angle, then it is equidistant from the two sides of the angle. If AD bisects BAC and DBAB and DCAC, then DB = DC THEOREM 5.5: CONVERSE OF THE ANGLE BISECTOR THEOREM If a point is in the interior of an angle and is equidistant from the sides of the angle, it lies on the angle bisector of the angle. If DBAB and DCAC and DB = DC, then AD is the of BAC. angle bisector

  11. Example 1 Use the Angle Bisector Thm Find the length of LM. JM bisects KJL because m K JM = m L JM. ML = KL ML = 5 Because JM bisects KJL and MK  JK and ML  JL Substitution Example 2 Use Algebra to solve a problem For what value of x does P lie on the bisector of GFH? P lies on the bisector of GFH if mGFP = mHFP. mGFP = mHFP 13x = 11x + 8 x = 4 Set angle measures equal. Substitute. Solve for x.

  12. Do the angle bisector part of the task Then show the geosketch example, showing how the center can move depending on the type of triangle used.

  13. THEOREM 5.7: CONCURRENCY OF ANGLE BISECTORS OF A TRIANGLE The angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle. If AP, BP, and CP are angle bisectors of ∆ ABC, then PD = The point of concurrency is called the incenter. Note: The incenter is always “inside” of the triangle. Note: The incenter is equal distance from all three sides. PE = PF

  14. Thm 5.7 Pythagorean Thm. Substitute known values. Multiply. Subtract 225 from both sides. Take Square Root of both sides. Substitute. VS = VT = VU a2 + b2 = c2 152 + VT2 + 172 225 + VT2 = 289 VT2 = 64 VT = 8 VS = 8

  15. NTG p.282 equidistant LI a2 + b2 c2 225 = LI2 + 144 152 LI2 + 122 – 144 = – 144 LI2 81 81 = LI2 LI 9 LI 9

  16. Solve for x. Solve for x. Because angles are congruent and the segments are perpendicular, then the segments are congruent. 10 = x + 3 x = 7 Because segments are congruent and perpendicular, then the angle is bisected which means they are are congruent. 9x – 1 = 6x + 14 3x = 15 x = 3 In the diagram, D is the incenter of ∆ABC. Find DF. DE = DF = DG DF = DG DF = 3 Concurrency of Angle Bisectors Substitution

  17. Assignment • Textbook: p266 (1-18) & p274 (1-12)

  18. 5.4 Use Medians and Altitudes TBK p.282 • Median of a triangle – a segment from a vertex to the midpoint of the opposite side • Centroid – point of concurrency of the three medians of a triangle (always inside the ) • Altitude of a triangle – perpendicular segment from a vertex to the opposite side or line that contains the opposite side (may have to extend the side of the triangle) • Orthocenter – point at which the lines containing the three altitudes of a triangle intersect Acute  = inside of  Right  = On the  Obtuse  = Outside of 

  19. Theorem 5.8 Concurrency of Medians of a Triangle (Centroid) The medians of a  intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side. In other words, the distance from the vertex to the centroid is twice the distance from the centroid to the midpoint. AB = AC + CB If AC is 2/3 of AB, what is CB? CB = 1/3 If AB = 9, what is AC and CB? AC = 6 What do you notice about AC and CB? AC is twice CB Why? Now assume, A is vertex, C is centroid, and B is midpoint of opposite side. Vertex to centroid = 2/3 median Centroid to midpoint = 1/3 median CB = 3

  20. P is the centroid of  ABC. What relationships exist? The dist. from the vertex to the centroid is twice the dist. from centroid to midpoint. AP = 2  PE BP = 2  PF CP = 2  PD The dist. from the centroid to midpoint is half the dist. from the vertex to the centroid. PE = ½  AP PF = ½  BP PD = ½  CP The dist. from the vertex to the centroid is 2/3 the distance of the median. AP = 2/3  AE BP = 2/3  BF CP = 2/3  CD The dist. from the centroid to midpoint is 1/3 the distance of the median. PE = 1/3  AE PF = 1/3  BF PD = 1/3  CD

  21. What do I know about DG? What do I know about BG? BG = BD + DG BG = 12 + 6 BG = 18 DG = 6

  22. Your Turn In PQR, S is the centroid, UQ = 5, TR = 3, RV = 5, and SU = 2. 1. Find RU and RS. RU is a median and RU = RS + SU. RS = 2  SU RU = RS + SU RU = 4 + 2 RS = 2  2 RU = 6 RS = 4 2. Find the perimeter of PQR. Perimeter means add up the sides of the triangle. U is midpoint of PQ so PU = UQ , PU = 5 QT, RU, and PV are medians since S is centroid. PQ = PU + UQ PQ = 10 RQ = RV + VQ RQ = 10 V is midpoint of RQ so RV = VQ, VQ = 5 T is midpoint of PR so RT = TP, TP = 3 RP = RT + TP RP = 6 Perimeter = PQ + QR + RP = 10 + 10 +6 = 26

  23. TBK p.278 Theorem 5.9 Concurrency of Altitudes of a Triangle (Orthocenter) The lines containing the altitudes of a triangle are concurrent. In a right triangle, the legs are also altitudes. In an obtuse triangle, sides of the triangle and/or the altitudes may have to be extended. Notice obtuse triangle, orthocenter is outside the triangle. Notice right triangle, orthocenter is on the triangle.

  24. Assignment • Textbook: p280-281 (1-6, 10-24)

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