Segments Proofs Reference

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1 Segments Proofs Reference Properties of Equality Addition Property Subtraction Property Multiplication Property Division Property Distributive Property Reflexive Property The properties above may only be used with EQUAL signs. The following properties of congruence can be applied to statements with congruence symbols: Properties of Congruence Reflexive Property of Congruence For any segment AB,. of Congruence of Congruence If, then. If and, then. Definitions Definition of Congruence Segments are congruence if and only if they have the same measure: If, then. If, then. Definition of Midpoint The midpoint of a segment divides the segment into 2 equal (congruent) parts. If M is the midpoint of AB, then Postulates If A, B, and C are collinear points and B is between A and C: Segment Addition Postulate A B C then:

2 Practice! Justify each of the following statements using a property of equality, property of congruence, definition, or postulate. 1. If PQ = PQ, then PQ PQ 2. If K is between J and L, then JK + KL = JL 3. EF EF 4. If RS = TU, then RS + XY = TU + XY 5. If AB = DE, then DE = AB 6. If Y is the midpoint of XZ, then XY = YZ 7. If FG HI and HI JK, then FG JK 8. If AB + CD = EF + CD, then AB = EF 9. If PQ + RS = TV and RS = WX, then PQ + WX = TV 10. If LP = PN, and L, P, and N are collinear, then P is the midpoint of LN 11. If UV UV, then UV = UV 12. If CD + DE = CE, then CD = CE DE Property Bank: Properties of Equality: Addition Property Subtraction Property Multiplication Property Division Property Distributive Property Reflexive Property Properties of Congruence: Reflexive Property Definitions: Postulates:

3 Segment Proofs Guide Directions: Use the reasons below to complete proofs Segment Proofs Guide Directions: Use the reasons below to complete proofs Addition Property Simplify Addition Property Simplify Substitution Substitution Definition of congruence Substitution Substitution Definition of congruence

4 Segments Proofs Directions: Complete the proofs below by giving the missing statements and reasons. 1 : E is the midpoint of DF Prove: 2DE = DF 1. E is the midpoint of DF DE = EF DE + DE = DE + EF DE = DE + EF DE + EF = DF DE = DF 6. D E F 2 : KL LN, LM LN Prove: L is the midpoint of KM K L N M 1. KL LN, LM LN KL = LN, LM = LN KL = LM L is the midpoint of KM 4. P U 3 : PQ TQ, UQ QS Prove: PS TU T Q S 1. PQ TQ, UQ QS PQ = TQ, UQ = QS PQ + QS = PS; TQ + QU = TU TQ + QS = PS TQ + QS = TU PS = TU PS TU 7.

5 J N 4 : K is the midpoint of JL, M is the midpoint of LN, JK = MN Prove: KL LM K L M 1. K is the midpoint of JL, M is the midpoint of LN JK = KL, LM = MN JK = MN MN = KL, LM = MN LM = KL KL = LM KL LM 7. 5 : XY UV, YZ TU Prove: XZ TV 1. XY UV, YZ TU XY = UV, YZ = TU XY + YZ = XZ, TU + UV = TV UV + YZ = XZ, YZ + UV = TV XZ = TV XZ TV 6. X Y Z T U V 6 : YW YZ, XY VY Prove: XZ VW V X Y Z W 1. WY YZ, XY VY WY = YZ, XY = VY XY + YZ = XZ VY + YW = XZ VY + YW = VW XZ = VW XZ VW 7.

6 7 : E is the midpoint of AC, DE = EC Prove: DE AE A E D C 1. E is the midpoint of AC AE = DE DE AE 6. 8 : RS = 2 1 RT Prove: S is the midpoint of RT R S T 1. RS = 2 1 RT RS = RT RS = RS + ST RS = ST : M is the midpoint of LN, N is the midpoint of MO Prove: LM NO L M N O 1. M is the midpoint of LN LM = MN MN = NO of Equality 6. 6.

7 10 : Y is the midpoint of XZ Prove: XY = 2 1 XZ 11 : AC DF, BC DE Prove: AB EF A B C D E F 12 : AB CD Prove: AC BD A B C D

8 Name: Date: Bell: Unit 2: Logic & Proof Homework 7: Segment Proofs ** This is a 2-page document! ** Use the segment addition postulate to write three equations using the diagram below P Q R S T 3. Complete the proofs below by filling in the missing statements and reasons Z 4. : X is the midpoint of WY, WX XZ Prove: XY XZ W X Y 1. X is the midpoint of WY WX = XY WX XZ WX = XZ XY = XZ XY XZ : AB CD Prove: AC BD A B C D 1. AB CD AB = CD AC + CD = AD AB + BD = AD CD + BD = AD AC + CD = CD + BD AC = BD AC BD 8.

9 6. : 2PQ = PR Prove: Q is the midpoint of PR P Q R PQ = PQ + QR PQ = QR : AB CD, BD DE A B C D Prove: AD CE E 1. AB CD, BD DE AB + BD = AD CD + DE = AD AD = CE : GI JL, GH KL Prove: HI JK G H I J K L

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