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Given: m CAT = m BAG Prove: m 1 = m 2

Given: m CAT = m BAG Prove: m 1 = m 2. Statements Reasons. 1. m CAT = m BAG 1. Given. 2. m 3 = m 3 2. Reflexive Property of Equality.

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Given: m CAT = m BAG Prove: m 1 = m 2

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  1. Given: m CAT = m BAG Prove: m 1 = m 2 Statements Reasons 1. m CAT = m BAG 1. Given 2. m 3 = m 3 2. Reflexive Property of Equality • m CAT = m 1 + m 3 3. Angle Addition Postulate • m BAG = m 2 + m 3 4. m 1 + m 3 = m 2 + m 3 4. Substitution Property 5. m 1 = m 2 5. Subtraction Property of Equality

  2. Given: ZP = LC; ZI = OC Prove: IP = LO Statements Reasons 1. ZP = LC; ZI = OC 1. Given 2. ZP = ZI + IP; LC = OC + LO 2. Segment Addition Postulate 3. ZI + IP = LO + OC 3. Substitution Property 4. IP = LO 4. Subtraction Property of Equality

  3. Given: m 1 = m 2; m 3 = m 4 Prove: m XYZ = m TUV Statements Reasons 1. m 1 = m 2; m 3 = m 4 1. Given • m 1 + m 3 = m XYZ 2. Angle Addition Postulate • m 2 + m 4 = m TUV 3. m 1 + m 3 = m 2 + m 4 3. Addition Property of Equality 4. m XYZ = m TUV 4. Substitution Property

  4. Given: MA = TH Prove: MT = AH Statements Reasons 1. MA = TH 1. Given 2. AT = AT 2. Reflexive Property of Equality • MT = MA + AT; 3. Segment Addition Postulate • AH = TH + AT 4. MA + AT = TH + AT 4. Addition Property of Equality 5. MT = AH 5. Substitution Property

  5. Given: is the bisector of BIT Prove: m BIG = m BIT; m GIT = m BIT Statements Reasons 1. is the bisector of BIT 1. Given 2. m BIG = m GIT 2. Definition of Angle Bisector 3. m BIG + m GIT = m BIT 3. Angle Addition Postulate 4. m BIG + m BIG = m BIT; 4. Substitution Property or 2m BIG = m BIT 5. m BIG = m BIT 5. Division Property of Equality 6. m GIT = m BIT 6. Substitution Property

  6. Given: LN = IE Prove: LI = NE Statements Reasons 1. LN = IE 1. Given 2. IN = IN 2. Reflexive Property of Equality 3. LN = LI + IN; IE = NE + IN 3. Segment Addition Postulate 4. LI + IN = NE + IN 4. Substitution Property 5. LI = NE 5. Subtraction Property of Equality

  7. Given: AE = BC; ED = CD Prove: AD = BD Statements Reasons 1. AE = BC; ED = CD 1. Given 2. AD = AE + ED; BD = BC + CD 2. Segment Addition Postulate 3. AE + ED = BC + CD 3. Addition Property of Equality 4. AD = BD 4. Substitution Property

  8. Given: m BAD = m ABD; m 1 = m 3 Prove: m 2 = m 4 Statements Reasons 1. m BAD = m ABD; m 1 = m 3 1. Given • m BAD = m 1 + m 2 2. Angle Addition Postulate • m ABD = m 3 + m 4 3. m 1 + m 2 = m 3 + m 4 3. Substitution Property 4. m 2 = m 4 4. Subtraction Property of Equality

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