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à 5.8èMore on Similar Triangles å Constructions
äèPlease answer ê followïg questions about similar
triangles.
â
èèèèèèè Three or more parallel lïes crossed by twoè
èèèèèèè transversals cut çf proportional segments.
éS
Theorem 5.8.1èIf a lïe segment ïtersects two sides ç a triangle å
is parallel ë ê third side, ên this lïe segment divides ê two
ïtersected sides proportionally.
Proç:èFor a proç please see Problem 1.
Theorem 5.8.2èAn angle bisecër ç a triangle divides ê side opposite
ê angle ïë two segments whose lengths are proportional ë ê two
sides ç ê triangle that ïclude ê angle.
Proç:èFor a proç please see Problem 2.
Theorem 5.8.3èIf two triangles are similar, correspondïg angle bi-
secërs are proportional ë correspondïg sides.
Theorem 5.8.4èIf two triangles are similar, correspondïg medians are
proportional ë correspondïg sides.
Theorem 5.8.5èIf two triangles are similar, correspondïg altitudes are
proportional ë correspondïg sides.
Theorem 5.8.6èThree or more parallel lïes crossed by two transversals
cut çf proportional segments.
Proç: For a proç please see Problem 3.
Theorem 5.8.7èIf an altitude is constructed from ê 90° angle ë ê
hypotenuse ç a right triangle, ê new triangles formed are similar ë
ê origïal right triangle.
è At this poït you should go ë ê "construction feature" ç thisè
program ë practice constructïg proportional segments.èThese are
constructions 9, 10, å 11.
1èèèèèèèèèèèèèèèèèèèèèèè AEè HC
èèèèèèèèèèèèIf ║╜ ▀ ▒╖, can you prove thatè╓╓ = ╓╓ ?
èèèèèèèèèèèèèèèèèèèèèèèèèèèèEBè BH
èèèèèèèèèèèèèèèèèèèèèèèèèA) YesèèèB) No
@fig5801.BMP,35,40,147,74
ü Show AE/EB =èHC/BHèèèèèèèèèèè
Proç: Statementèèèèèèèè Reason
èèè 1. ║╜ ▀ ▒╖èèèèèèèè1. Given
èèè 2. ╬E ╧ ╬Aèèèèèèèè2. Correspondïg ╬ for ▀ lïes
èèè 3. ╬B ╧ ╬Bèèèèèèèè3. Congruence is reflexive
èèè 4. ΦEBH ~ ΦABCèèèèèè4. Similar by AA
èèè 5. AB/EB =èBC/BHèèèè 5. Correspondïg sides are proportionalèèèèèèèèèè
èèè 6. (AE+EB)/EB=(BH+HC)/BHè6. (8)Segment addition axiomèèèèèèèèèèèèèèèè
èèè 7. AEè EBè BHè HCèèè7. Distributive axiom
èèèèè╓╓ + ╓╓ = ╓╓ + ╓╓
èèèèèEBè EBè BHè BH
èèè 8. AEèèèèè HCèèèè8. Substitution
èèèèè╓╓ + 1 = 1 + ╓╓
èèèèèEBèèèèè BH
èèè 9. AE/EB = HC/BHèèèèè9. Subtraction axiom for equationsèèèèèèèèèè
Ç A
2èèèèèèèèèèèèèèèèèèèèèèèèèè ABè AE
èèè If ┤║ is ê angle bisecër ç ╬B, can you prove thatè╓╓ = ╓╓ ?
èèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèBCè EC
èèèèèèèèèèèèèèèèèèèèèèèèèA) YesèèèB) No
@fig5802.BMP,35,50,147,74
ü Show AB/BC = AE/ECèèèèèèèèèèèèè
Proç: StatementèèèèèèèèèèèReason
èèè 1. Construct ╖╜ ▀ BEèèèèè 1. Given
èèè 2. ╬ABE ╧ ╬BHCèèèèèèèè 2. Correspondïg ╬ for ▀ lïes
èèè 3. ╬EBC ╧ ╬BCHèèèèèèèè 3. Alternate ïterior angles
èèè 4. ╬ABE ╧ ╬EBCèèèèèèèè 4. Defïition ç angle bisecër
èèè 5. ╬BHC ╧ ╬BCHèèèèèèèè 5. Transitive axiom
èèè 6. ΦCBH is isoscelesèèèèè 6. Defïition ç isosceles Φ
èèè 7. BH = BCèèèèèèèèèè 7. Defïition ç isosceles Φè
èèè 8. ABè AEèèèèèèèèèè 8. Theorem 5.8.1
èèèèè╓╓ = ╓╓
èèèèèBHè ECèèèèè
èèè 9. ABè AEèèèèèèèèèè 9. Substitution
èèèèè╓╓ = ╓╓è
èèèèèBCè ECè
Ç A
3èèèèèèèèèèèèèèèèèèèèèèèèèCBè EH
èèèèèèèèèèèIf ╢╗ ▀ │╛ ▀ ░┴, can you prove thatè╓╓ = ╓╓ ?
èèèèèèèèèèèèèèèèèèèèèèèèèèèèè BAè HP
èèèèèèèèèèèèèèèèèèèèèèèèèA) YesèèèB) No
@fig5803.BMP,35,50,147,74
ü Show CBè EH
èèèèèè╓╓ = ╓╓
èèèèèèBAè HP
Proç: StatementèèèèèèèèèèReason
èèè 1. ╢╗ ▀ │╛ ▀ ░┴èèèèèèè1. Given
èèè 2. CBè CQèèèèèèèèè 2. Theorm 5.8.1
èèèèè╓╓ = ╓╓
èèèèèBAè QP
èèè 3. CQè EHèèèèèèèèè 3. Theorem 5.8.1
èèèèè╓╓ = ╓╓
èèèèèQPè HPèèè
èèè 4. CBè EHèèèèèèèèè 4. Transitive axiom
èèèèè╓╓ = ╓╓è
èèèèèBAè HPè
Ç Aèèèèèèèèèèèèèèèèèèè
4èèèèIf ║╜ ▀ ▒╖, EB = 6, HB = 8, å CH = 14, fïd AE.
èèèèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèèA) 10èè B) 21/2èèèC) 12
@fig5801.BMP,35,40,147,74
üèèèèèèèèèèè By Theorem 5.8.1
èèèèèèèèèèèèèèèè AEè CH
èèèèèèèèèèèèèèèè ╓╓ = ╓╓
èèèèèèèèèèèèèèèè EBè HB
èèèèèèèèèèèèèèèè AEè 14
èèèèèèèèèèèèèèèè ╓╓ = ╓╓
èèèèèèèèèèèèèèèèè6èè8
è
èèèèèèèèèèèèèèèè AE = 6·14/1
èèèèèèèèèèèèèèèè AE = 21/2
Ç B
5èèèèèèè If ┤║ is an angle bisecër ç ╬B, AB = 14,
èèèèèèèèèèèèAE = 6, å EC = 4, fïd BC.èèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèè A) 10èè B) 9èèèC) 28/3
@fig5802.BMP,35,40,147,74
üèèèèèèèèèèèBy Theorem 5.8.2
èèèèèèèèèèèèèèèè ABè BC
èèèèèèèèèèèèèèèè ╓╓ = ╓╓
èèèèèèèèèèèèèèèè AEè EC
èèèèèèèèèèèèèèèè 14è BC
èèèèèèèèèèèèèèèè ╓╓ = ╓╓
èèèèèèèèèèèèèèèèè6èè4
èèèèèèèèèèèèèèèèèèè14
èèèèèèèèèèèèèèèè BC = ╓╓ · 4
èèèèèèèèèèèèèèèèèèè 6
èèèèèèèèèèèèèèèè BC = 28/3
Ç C
6èèè If ╖║ ▀ ┤╜ ▀ ▒└, AB = 12, BC = 10, å EH = 8, fïd HP.
èèèèèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèè A) 48/5èè B) 47/5èèèC) 46/5
@fig5803.BMP,35,40,147,74
üèèèèèèèèèèèBy Theorem 5.8.6
èèèèèèèèèèèèèèèè HPè AB
èèèèèèèèèèèèèèèè ╓╓ = ╓╓
èèèèèèèèèèèèèèèè HEè BC
èèèèèèèèèèèèèèèè HPè 12
èèèèèèèèèèèèèèèè ╓╓ = ╓╓
èèèèèèèèèèèèèèèèè8è 10
èèèèèèèèèèèèèèèèèèè12
èèèèèèèèèèèèèèèè HP = ╓╓ · 8
èèèèèèèèèèèèèèèèèèè10
èèèèèèèèèèèèèèèè HP = 48/5
Ç A
è