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5.1 midsegments of triangles Geometry Mrs. Spitz Fall 2004.

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Presentation on theme: "5.1 midsegments of triangles Geometry Mrs. Spitz Fall 2004."— Presentation transcript:

1 5.1 midsegments of triangles Geometry Mrs. Spitz Fall 2004

2 Objectives: Use properties of midsegments to solve problems Use properties of perpendicular bisectors Use properties of angle bisectors to identify equal distances such as the lengths of beams in a room truss.

3 Midsegments of triangles Turn in your book to page 243. In the blue box is an investigation. Complete the investigation, we’ll share our conjectures in 3 minutes.

4 Midsegment A segment that connects the midpoints of two sides Midsegment theorem: if a segment joins the midpoints of two sides of a triangle, then the segment is parallel to the third side AND is half it’s length.

5 Example 1 In triangle XYZ, M, N, and P are midpoints. The perimeter of triangle MNP is 60. Find NP and YZ. Y X Z M P N 22 24

6 Example 2 In triangle DEF, A, B, and C are midpoints. Name pairs of parallel segments.

7 Check for understanding. –AB= 10 and CD = 18. Find EB, BC, and AC. Critical Thinking –Find m<VUZ. Justify your answer D A E B C = = | | V X Y U Z = = ||

8 Real world connection CD is a new bridge being built over a lake as shown. Find the length of the bridge.

9 Assignment: Page 246 #1-20 we’ll go over them then you’ll turn in tomorrow #’s 21-36; 47-55

10 5.2 Bisectors in Triangles

11 Objectives: Use properties of midsegments to solve problems Use properties of perpendicular bisectors Use properties of angle bisectors to identify equal distances such as the lengths of beams in a room truss.

12 Use Properties of perpendicular bisectors In lesson 1.5, you learned that a segment bisector intersects a segment at its midpoint. A segment, ray, line, or plane that is perpendicular to a segment at its midpoint is called a perpendicular bisector. CP is a  bisector of AB

13 Equidistant A point is equidistant from two points if its distance from each point is the same. In the construction above, C is equidistant from A and B because C was drawn so that CA = CB.

14 Perpendicular bisector theorem If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.

15 Theorem 5.1 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 perpendicular bisector of AB, then CA = CB.

16 Converse of the perpendicular bisector theorem If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

17 Theorem 5.2: Converse of the Perpendicular Bisector Theorem If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment. If DA = DB, then D lies on the perpendicular bisector of AB.

18 Plan for Proof of Theorem 5.1 Refer to the diagram for Theorem 5.1. Suppose that you are given that CP is the perpendicular bisector of AB. Show that right triangles ∆ABC and ∆BPC are congruent using the SAS Congruence Postulate. Then show that CA ≅ CB.

19 Statements: 1.CP is perpendicular bisector of AB. 2.CP  AB 3.AP ≅ BP 4. CP ≅ CP 5.  CPB ≅  CPA 6.∆APC ≅ ∆BPC 7.CA ≅ CB Reasons: 1.Given Given: CP is perpendicular to AB. Prove: CA ≅CB

20 Statements: 1.CP is perpendicular bisector of AB. 2.CP  AB 3.AP ≅ BP 4. CP ≅ CP 5.  CPB ≅  CPA 6.∆APC ≅ ∆BPC 7.CA ≅ CB Reasons: 1.Given 2.Definition of Perpendicular bisector Given: CP is perpendicular to AB. Prove: CA ≅CB

21 Statements: 1.CP is perpendicular bisector of AB. 2.CP  AB 3.AP ≅ BP 4. CP ≅ CP 5.  CPB ≅  CPA 6.∆APC ≅ ∆BPC 7.CA ≅ CB Reasons: 1.Given 2.Definition of Perpendicular bisector 3.Given Given: CP is perpendicular to AB. Prove: CA ≅CB

22 Statements: 1.CP is perpendicular bisector of AB. 2.CP  AB 3.AP ≅ BP 4. CP ≅ CP 5.  CPB ≅  CPA 6.∆APC ≅ ∆BPC 7.CA ≅ CB Reasons: 1.Given 2.Definition of Perpendicular bisector 3.Given 4.Reflexive Prop. Congruence. Given: CP is perpendicular to AB. Prove: CA ≅CB

23 Statements: 1.CP is perpendicular bisector of AB. 2.CP  AB 3.AP ≅ BP 4. CP ≅ CP 5.  CPB ≅  CPA 6.∆APC ≅ ∆BPC 7.CA ≅ CB Reasons: 1.Given 2.Definition of Perpendicular bisector 3.Given 4.Reflexive Prop. Congruence. 5.Definition right angle Given: CP is perpendicular to AB. Prove: CA ≅CB

24 Statements: 1.CP is perpendicular bisector of AB. 2.CP  AB 3.AP ≅ BP 4. CP ≅ CP 5.  CPB ≅  CPA 6.∆APC ≅ ∆BPC 7.CA ≅ CB Reasons: 1.Given 2.Definition of Perpendicular bisector 3.Given 4.Reflexive Prop. Congruence. 5.Definition right angle 6.SAS Congruence Given: CP is perpendicular to AB. Prove: CA ≅CB

25 Statements: 1.CP is perpendicular bisector of AB. 2.CP  AB 3.AP ≅ BP 4. CP ≅ CP 5.  CPB ≅  CPA 6.∆APC ≅ ∆BPC 7.CA ≅ CB Reasons: 1.Given 2.Definition of Perpendicular bisector 3.Given 4.Reflexive Prop. Congruence. 5.Definition right angle 6.SAS Congruence 7.CPCTC Given: CP is perpendicular to AB. Prove: CA ≅CB

26 Ex. 1 Using Perpendicular Bisectors In the diagram MN is the perpendicular bisector of ST. a.What segment lengths in the diagram are equal? b.Explain why Q is on MN.

27 Ex. 1 Using Perpendicular Bisectors a.What segment lengths in the diagram are equal? Solution: MN bisects ST, so NS = NT. Because M is on the perpendicular bisector of ST, MS = MT. (By Theorem 5.1). The diagram shows that QS = QT = 12.

28 Ex. 1 Using Perpendicular Bisectors b.Explain why Q is on MN. Solution: QS = QT, so Q is equidistant from S and T. By Theorem 5.2, Q is on the perpendicular bisector of ST, which is MN.

29 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 m  BAD = m  CAD, then DB = DC

30 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, then it lies on the bisector of the angle. If DB = DC, then m  BAD = m  CAD.

31 Ex. 2: Proof of angle bisector theorem Given: D is on the bisector of  BAC. DB  AB, DC  AC. Prove: DB = DC Plan for Proof: Prove that ∆ADB ≅ ∆ADC. Then conclude that DB ≅DC, so DB = DC.

32 Paragraph Proof By definition of an angle bisector,  BAD ≅  CAD. Because  ABD and  ACD are right angles,  ABD ≅  ACD. By the Reflexive Property of Congruence, AD ≅ AD. Then ∆ADB ≅ ∆ADC by the AAS Congruence Theorem. By CPCTC, DB ≅ DC. By the definition of congruent segments DB = DC.

33 Developing Proof Given: D is in the interior of  ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of  ABC.

34 Statements: 1.D is in the interior of  ABC. 2.D is ___?_ from BA and BC. 3.____ = ____ 4.DA  ____, ____  BC 5.__________ 6.__________ 7.BD ≅ BD 8. __________ 9.  ABD ≅  CBD 10. BD bisects  ABC and point D is on the bisector of  ABC Reasons: 1.Given Given: D is in the interior of  ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of  ABC.

35 Statements: 1.D is in the interior of  ABC. 2.D is EQUIDISTANT from BA and BC. 3.____ = ____ 4.DA  ____, ____  BC 5.__________ 6.__________ 7.BD ≅ BD 8. __________ 9.  ABD ≅  CBD 10. BD bisects  ABC and point D is on the bisector of  ABC Reasons: 1.Given 2.Given Given: D is in the interior of  ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of  ABC.

36 Statements: 1.D is in the interior of  ABC. 2.D is EQUIDISTANT from BA and BC. 3.DA = DC 4.DA  ____, ____  BC 5.__________ 6.__________ 7.BD ≅ BD 8. __________ 9.  ABD ≅  CBD 10. BD bisects  ABC and point D is on the bisector of  ABC Reasons: 1.Given 2.Given 3.Def. Equidistant Given: D is in the interior of  ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of  ABC.

37 Statements: 1.D is in the interior of  ABC. 2.D is EQUIDISTANT from BA and BC. 3.DA = DC 4.DA  _BA_, __DC_  BC 5.__________ 6.__________ 7.BD ≅ BD 8. __________ 9.  ABD ≅  CBD 10. BD bisects  ABC and point D is on the bisector of  ABC Reasons: 1.Given 2.Given 3.Def. Equidistant 4.Def. Distance from point to line. Given: D is in the interior of  ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of  ABC.

38 Statements: 1.D is in the interior of  ABC. 2.D is EQUIDISTANT from BA and BC. 3.DA = DC 4.DA  _BA_, __DC_  BC 5.  DAB = 90 °  DCB = 90 ° 6.__________ 7.BD ≅ BD 8. __________ 9.  ABD ≅  CBD 10. BD bisects  ABC and point D is on the bisector of  ABC Reasons: 1.Given 2.Given 3.Def. Equidistant 4.Def. Distance from point to line. 5.If 2 lines are , then they form 4 rt.  s. Given: D is in the interior of  ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of  ABC.

39 Statements: 1.D is in the interior of  ABC. 2.D is EQUIDISTANT from BA and BC. 3.DA = DC 4.DA  _BA_, __DC_  BC 5.  DAB and  DCB are rt.  s 6.  DAB = 90 °  DCB = 90 ° 7.BD ≅ BD 8. __________ 9.  ABD ≅  CBD 10. BD bisects  ABC and point D is on the bisector of  ABC Reasons: 1.Given 2.Given 3.Def. Equidistant 4.Def. Distance from point to line. 5.If 2 lines are , then they form 4 rt.  s. 6.Def. of a Right Angle Given: D is in the interior of  ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of  ABC.

40 Statements: 1.D is in the interior of  ABC. 2.D is EQUIDISTANT from BA and BC. 3.DA = DC 4.DA  _BA_, __DC_  BC 5.  DAB and  DCB are rt.  s 6.  DAB = 90 °  DCB = 90 ° 7.BD ≅ BD 8. __________ 9.  ABD ≅  CBD 10. BD bisects  ABC and point D is on the bisector of  ABC Reasons: 1.Given 2.Given 3.Def. Equidistant 4.Def. Distance from point to line. 5.If 2 lines are , then they form 4 rt.  s. 6.Def. of a Right Angle 7.Reflexive Property of Cong. Given: D is in the interior of  ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of  ABC.

41 Statements: 1.D is in the interior of  ABC. 2.D is EQUIDISTANT from BA and BC. 3.DA = DC 4.DA  _BA_, __DC_  BC 5.  DAB and  DCB are rt.  s 6.  DAB = 90 °  DCB = 90 ° 7.BD ≅ BD 8.∆ABD ≅ ∆CBD 9.  ABD ≅  CBD 10. BD bisects  ABC and point D is on the bisector of  ABC Reasons: 1.Given 2.Given 3.Def. Equidistant 4.Def. Distance from point to line. 5.If 2 lines are , then they form 4 rt.  s. 6.Def. of a Right Angle 7.Reflexive Property of Cong. 8.HL Congruence Thm. Given: D is in the interior of  ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of  ABC.

42 Statements: 1.D is in the interior of  ABC. 2.D is EQUIDISTANT from BA and BC. 3.DA = DC 4.DA  _BA_, __DC_  BC 5.  DAB and  DCB are rt.  s 6.  DAB = 90 °  DCB = 90 ° 7.BD ≅ BD 8.∆ABD ≅ ∆CBD 9.  ABD ≅  CBD 10. BD bisects  ABC and point D is on the bisector of  ABC Reasons: 1.Given 2.Given 3.Def. Equidistant 4.Def. Distance from point to line. 5.If 2 lines are , then they form 4 rt.  s. 6.Def. of a Right Angle 7.Reflexive Property of Cong. 8.HL Congruence Thm. 9.CPCTC Given: D is in the interior of  ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of  ABC.

43 Statements: 1.D is in the interior of  ABC. 2.D is EQUIDISTANT from BA and BC. 3.DA = DC 4.DA  _BA_, __DC_  BC 5.  DAB and  DCB are rt.  s 6.  DAB = 90 °  DCB = 90 ° 7.BD ≅ BD 8.∆ABD ≅ ∆CBD 9.  ABD ≅  CBD 10. BD bisects  ABC and point D is on the bisector of  ABC Reasons: 1.Given 2.Given 3.Def. Equidistant 4.Def. Distance from point to line. 5.If 2 lines are , then they form 4 rt.  s. 6.Def. of a Right Angle 7.Reflexive Property of Cong. 8.HL Congruence Thm. 9.CPCTC 10.Angle Bisector Thm. Given: D is in the interior of  ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of  ABC.

44 Assignment: You’ll do page 251 #1-11 we’ll go over them then you’ll turn #’s 12-31; 41, 42, 55- 63


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