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GEOMETRY
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CHAPTER3.5Y
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à 3.5èTheorems Involvïg ê Angles ç a Triangle
äèPlease answer ê followïg questions about ê angles
ç triangles.
â
èèèèThe sum ç ê ïterior angles ç a triangle is 180°.
éS1èèèèèèèèèèèèIn Section 3.2 we defïed remote å
èèèèèèèèèèèèèèèèexterior angles ç a triangle, å ïèèèèèèè
èèèèèèèèèèèèèèèèSection 3.4 we looked at alternate ï-
èèèèèèèèèèèèèèèèterior angles for parallel lïes. In
èèèèèèèèèèèèèèèèthis section we will combïe this ï-
èèèèèèèèèèèèèèèèformation å gaï some additional facts
@fig3501.BMP,40,40,147,74èèè about ê angles ç a triangle.
Theorem 3.5.1èThe sum ç ê measures ç ê ïterior angles ç a tri-
angle is 180°.
Proç: Statementèèèèèèèèè Reason
èèè 1. Construct ╣╛ ▀ ░╕èèèè1. Problem 5, Section 3.4
èèèèèèthrough Bè
èèè 2. m╬EBC + m╬3 = 180°èèè 2. (15)Lïear pairs are supplementary
èèè 3. m╬EBC = m╬1 + m╬2èèèè3. (12)Angle addition axiom
èèè 4. m╬1 + m╬2 + m╬3 = 180°è 4. Substitution
èèè 5. ╬1 ╧ A, ╬3 ╧ ╬Cèèèèè5. Alternate ïterior angles are
èèèèèèèèèèèèèèèèèèèècongruent
èèè 6. m╬1 = m╬A, m╬3 = m╬Cèè 6. Defïition ç congruence
èèè 7. m╬A + m╬2 + m╬C = 180°è 7. Substitution
Theorem 3.5.2èThe measure ç an exterior angle ç a triangle is equal
ë ê sum ç ê measures ç ê correspondïg remote ïterior angles.
Proç: Please see Problem 1 for ê proç.
Theorem 3.5.3èThe sum ç ê measures ç ê exterior angles ç a
triangle is 720°.
Proç: Please see Problem 2 for ê proç.
Theorem 3.5.4èIf two angles ç one triangle are congruent ë two angles
ç anoêr triangle, ên ê remaïïg correspondïg angles are con-
gruent.
Proç: Please see problem 3 for proç.
Theorem 3.5.5èIf two angles å ê side opposite one ç ê angles ï
one triangle are congruent ë ê correspondïg parts ç anoêr tri-
angle, ên ê triangles are congruent.
Proç: Please see Problem 4 for ê proç.
Theorem 3.5.6èCongruence ç triangles is reflexive, symmetric, å
transitive.
Proç: This follows from ê reflexive, symmetric, å transitive
properties ç angle å segment congruence.
1èèèèèèèè Can it be shown that m╬BCE = m╬A + m╬B?è
@fig3201.BMP,35,40,147,74èèèèè A) YesèèèB) No
ü Show m╬BCE = m╬A + m╬B
Proç: Statementèèèèèèèèèè Reason
èèè 1. m╬A + m╬B + m╬ACB = 180°è 1. Sum ç ïterior ╬s is 180°
èèè 2. m╬ACB + m╬BCE = 180°èèè 2. Lïear pairs are supplementary
èèè 3. m╬A + m╬B + m╬ACB =èèèè3. Transitive axiom
èèèèèèèèèè m╬ACB + m╬BCE
èèè 4. m╬A + m╬B = m╬BCEèèèèè4. Subtraction axiom for equations
Ç A
2èèèèèèèèèè Can it be shown that ê sum ç all ç êèè
èèèèèèèèèèèèèèèexterior angles ç this triangle is 360°?
@fig3202.BMP,35,40,147,74èèèèè A) YesèèèB) No
ü Show ê sum is not 360°
Proç: Statementèèèèèèè Reason
èèè 1. m╬A + m╬B = m╬BCRèè1. Exterior ╬s equal sum ç two remote ╬s
èèè 2. m╬A + m╬B = m╬ACSèè2. Exterior ╬s equal sum ç two remote ╬s
èèè 3. m╬B + m╬C = m╬BAHèè3. Exterior ╬s equal sum ç two remote ╬s
èèè 4. m╬B + m╬C = m╬EACèè4. Exterior ╬s equal sum ç two remote ╬s
èèè 5. m╬A + m╬C = m╬PBAèè5. Exterior ╬s equal sum ç two remote ╬s
èèè 6. m╬A + m╬C = m╬QBCèè6. Exterior ╬s equal sum ç two remote ╬s
èèè 7. 4(m╬A + m╬B + m╬C)è 7. Addition axiom for equations
èèèèè= sum ç exterior ╬s
èèè 8. 4(180) = sum çèèè8. Sum ç ïterior ╬s is 180°
èèèèèexterior ╬s
èèè 9. 720 = sum çèèèè 9. Multiplication is closed
èèèèèexterior ╬s
Ç B
3è Can it be shown that if ╬C ╧ ╬E å ╬B ╧ ╬H, ên ╬A ╧ ╬P?è
èèèèèèèèèèèèèèèèèèèèè
@fig3502.BMP,35,40,147,74èèèèè A) YesèèèB) No
ü Show ╬A ╧ ╬P
Proç: StatementèèèèèèèèèèèReason
èèè 1. ╬C ╧ ╬Eèèèèèèèèèè 1. Given
èèè 2. ╬B ╧ ╬Hèèèèèèèèèè 2. Given
èèè 3. m╬C = m╬Eèèèèèèèèè 3. Defïition ç congruence
èèè 4. m╬B = m╬Hèèèèèèèèè 4. Defïition ç congruence
èèè 5. m╬A + m╬B + m╬C = 180°èèè5. Sum ç ïterior ╬s is 180°
èèè 6. m╬P + m╬H + m╬E = 180°èèè6. Sum ç ïterior ╬s is 180°
èèè 7. m╬A + m╬B + m╬C =èèèèè 7. Transitive axiom
èèèèèèm╬P + m╬H + m╬E
èèè 8. m╬A = m╬Pèèèèèèèèè 8. Substitution å subtractionè
èèèèèèèèèèèèèèèèèèèèè axioms for equations
èèè 9. ╬A ╧ ╬Pèèèèèèèèèè 9. Defïition ç congruenceè
Ç A
4è Can it be shown that if ╬A ╧ ╬P, ╬C ╧ ╬E, å ▒┤ ╧ └╜,èè
èèèèèèèèèèèèèèèèèèèèèèèèên ΦABC ╧ ΦPHE?
@fig3502.BMP,35,40,147,74èèèèè A) YesèèèB) No
ü Show ΦABC ╧ ΦPHE
Proç: StatementèèèèèèèèèèèReason
èèè 1. ╬A ╧ ╬Pèèèèèèèèèè 1. Given
èèè 2. ╬C ╧ ╬Eèèèèèèèèèè 2. Given
èèè 3. ▒┤ ╧ └╜èèèèèèèèèè 3. Given
èèè 4. ╬B = ╬Hèèèèèèèèèè 4. Problem 3 above
èèè 5. ΦABC ╧ ΦPHEèèèèèèèè 5. Congruent by ASA
Ç Aèèè
5è Can it be shown that ï a right triangle, ê two acuteèè
èèèèèèèèèèèèèèèèèèèèèangles are complementary?
@fig3102.BMP,35,40,147,74èèèèè A) YesèèèB) No
ü Show ╬A å ╬B are complementary
Proç: StatementèèèèèèèèèèèReason
èèè 1. m╬A + m╬B + m╬C = 180°èèè1. Sum ç ïterior ╬s is 180°
èèè 2. m╬C = 90°èèèèèèèèè 2. Given
èèè 3. m╬A + m╬B + 90° = 180°èèè3. Substitution
èèè 4. m╬A + m╬B = 90°èèèèèè 4. Subtraction axiom for equations
èèè 5. ╬A å ╬B areèèèèèèè 5. Defïition ç complementary
èèèèè complementaryèèèèèèèèè angles
Ç Aèèè
6è Can it be shown that ï an equiangular triangle, each angleèè
èèèèèèèèèèèèèèèèèèèèèèèèhas ê measure ç 60°?
@fig3104.BMP,35,40,147,74èèèèè A) YesèèèB) No
ü Show m╬A = m╬B = m╬C = 60°
Proç: StatementèèèèèèèèèèReason
èèè 1. ΦABC is equiangularèèè 1. Given
èèè 2. m╬A = m╬B = m╬Cèèèèè 2. Defïition ç equiangularè
èèè 3. m╬A + m╬B + m╬C = 180°èè3. Sum ç ïterior ╬s is 180°
èèè 4. m╬A + m╬A + m╬A = 180°èè4. Substitution
èèè 5. 3(m╬A) = 180°èèèèèè 5. Distributive axiom
èèè 6. m╬A = 60°èèèèèèèè 6. Multiplication axiom ç equations
èèè 7. m╬B = 60° å m╬C = 60°è 7. Substitutionèè
Ç Aèèè
7èè Can it be shown that ê right triangles are congruentèèè
èèèèèèèèèèèèèèèèèèèèèèèèif ▒╖ ╧ └║, ┤╖ ╧ ╜║?
@fig3503.BMP,35,40,147,74èèèèè A) YesèèèB) No
ü Show ΦABC ╧ ΦPHE
Proç: StatementèèèèèèèèèReason
èèè 1. ▒╖ ╧ └║èèèèèèèè 1. Given
èèè 2. ╖┤ ╧ ║╜èèèèèèèè 2. Givenè
èèè 3. ╬C ╧ ╬Eèèèèèèèè 3. (14)All right angles are congruent
èèè 4. ΦABC ╧ ΦPHEèèèèèè 4. Congruent by SASèèèèèèèèèè
Ç Aèèè
8èèè Can it be shown that ê right triangles are congruentèèè
èèèèèèèèèèèèèèèèèèèèèèè if ▒┤ ╧ └╜ å ╬A ╧ ╬P?
@fig3503.BMP,35,40,147,74èèèèè A) YesèèèB) No
ü Show ΦABC ╧ ΦPHE
Proç: StatementèèèèèèèèèReason
èèè 1. ▒┤ ╧ └╜èèèèèèèè 1. Given
èèè 2. ╬A ╧ ╬Pèèèèèèèè 2. Givenè
èèè 3. ╬C ╧ ╬Eèèèèèèèè 3. (14)All right angles are congruent
èèè 4. ΦABC ╧ ΦPHEèèèèèè 4. Congruent by Theorem 3.5.5èèèèèèèèèè
Ç Aèèè
9èè Can it be shown that ê right triangles are congruentèèè
èèèèèèèèèèèèèèèèèèèèèè if ┤╖ ╧ ╜║ å ╬A ╧ ╬P?
@fig3503.BMP,35,40,147,74èèèèè A) YesèèèB) No
ü Show ΦABC ╧ ΦPHE
Proç: StatementèèèèèèèèèReason
èèè 1. ┤╖ ╧ ╜║èèèèèèèè 1. Given
èèè 2. ╬A ╧ ╬Pèèèèèèèè 2. Givenè
èèè 3. ╬C ╧ ╬Eèèèèèèèè 3. (14)All right angles are congruent
èèè 4. ╬B ╧ ╬Hèèèèèèèè 4. Problem 3 above
èèè 5. ΦABC ╧ ΦPHEèèèèèè 4. Congruent by ASAèèèèèèèèèè
Ç Aèèè
10èèè Can it be shown that ΦABC ╧ ΦEHC if segments ▒║ å ┤╜èèèè
èèèèèèèèèèèèèèè are perpendicular å bisect each oêr?
@fig3504.BMP,35,40,147,74èèèèè A) YesèèèB) No
ü Show ΦABC ╧ ΦEHC
Proç: Statementèèèèèèèèèèè Reason
èèè 1. ▒║ ß ┤╜èèèèèèèèèèè1. Given
èèè 2. ╬ACB, ╬ECH are right ╬sèèè2. Defïition ç perpendicularè
èèè 3. ╬ACE ╧ ╬ECHèèèèèèèèè3. (14)All right angles are
èèèèèèèèèèèèèèèèèèèèèècongruent
èèè 4. ▒╖ bisects ┤╜ åèèèèèè4. Given
èèèèèè┤╜ bisects ▒║
èèè 5. AC = CE, BC = CHèèèèèè 5. Defïition ç bisect
èèè 6. ▒╖ ╧ ╖║, ┤╖ ╧ ╖╜èèèèèè 6. Defïition ç congruence
èèè 7. ΦABC ╧ ΦEHCèèèèèèèèè7. Congruent by SASèèèèèè
Ç Aèèè
èèè
èèèè
èèèèèèèèè
èèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèè