دليل معلم مقرر رياضيات 1

 3 :‫ﺍﻟﻨﺸﺎﻁ ﺍﻟﺴﺎﺑﻖ ﻳﺒ ﹼﻴﻦ ﺃﺭﺑﻊ ﻃﺮﺍﺋﻖ ﻹﺛﺒﺎﺕ ﺗﻄﺎﺑﻖ ﺍﻟﻤﺜﻠﺜﺎﺕ ﺍﻟﻘﺎﺋﻤﺔ ﻭﻫﻲ‬  ✓      ‫؛ ﻟﺘﻘﻮﻳﻢ ﻓﻬﻢ ﺍﻟﻄﻼﺏ‬10 ‫ﺍﺳﺘﻌﻤﻞ ﺍﻟﺴﺆﺍﻝ‬ ‫ﻃﺮﻳﻘﺔ ﻛﺘﺎﺑﺔ ﺑﺮﻫﺎﻥ ﻧﻈﺮﻳﺔ ﺑﺎﻻﺳﺘﻨﺎﺩ‬  .‫ﻟﻤﺴﻠﻤﺎﺕ‬ LL 3.6          leg L H  ‫ﺍﻃﻠﺐ ﺇﻟﻰ ﺍﻟﻄﻼﺏ ﻛﺘﺎﺑﺔ ﺑﺮﻫﺎ ﹴﻥ ﺣ ﱟﺮ؛ ﻹﺛﺒﺎﺕ‬  .‫ ﻓﻲ ﺍﻟﻤﺜﻠﺜﺎﺕ ﺍﻟﻘﺎﺋﻤﺔ‬LL ‫ﺣﺎﻟﺔ ﺍﻟﺘﻄﺎﺑﻖ‬ Hypotenuse HA 3.7 Angle A DEF , RST      D    R A Y X L A 3.8 EF C B Z  ST      . ‫ﻣﺜﻠﺜﺎﻥ ﻗﺎﺋﻤﺎ ﺍﻟﺰﺍﻭﻳﺔ‬  . ‫∠ ﻗﺎﺋﻤﺘﺎﻥ‬E , ∠S HL 3.9 EF ST , ED SR        DEF RST   EF ST , ED :‫ﻧﻌﻠﻢ ﺃﻥ‬  ‫ ﻭﺑﻤﺎ ﺃﻥ ﺟﻤﻴﻊ‬،‫∠ ﻗﺎﺋﻤﺘﺎﻥ‬E , ∠S ‫ ﻭﺃﻥ‬، SR ،∠E ∠S ‫ ﺇﺫﻥ‬،‫ﺍﻟﺰﻭﺍﻳﺎ ﺍﻟﻘﺎﺋﻤﺔ ﻣﺘﻄﺎﺑﻘﺔ‬ :‫ ﻓﺎﺫﻛﺮ ﺍﻟﻤﺴﻠﻤﺔ ﺃﻭ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺘﻲ ﺍﺳﺘﻌﻤﻠﺘﻬﺎ‬،“‫ ﻭﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻹﺟﺎﺑﺔ ”ﻧﻌﻢ‬،‫ﺣ ﹼﺪﺩ ﻣﺎ ﺇﺫﺍ ﻛﺎﻥ ﻛﻞ ﺯﻭﺝ ﻣﻦ ﺍﻟﻤﺜﻠﺜﺎﺕ ﺍﻵﺗﻴﺔ ﻣﺘﻄﺎﺑﻘﻴﻦ ﺃﻡ ﻻ‬ ‫ ﺑﺤﺴﺐ‬DEF RST ‫ﻭﻟﺬﻟﻚ ﻳﻜﻮﻥ‬ HL ،‫ﻧﻌﻢ‬ (9 ‫ﻻ‬ (8 LA ،‫ﻧﻌﻢ‬ (7 . SAS ‫ ﺍﺳﺘﻌﻤﻞ ﻧﻈﺮﻳﺔ ﻓﻴﺜﺎﻏﻮﺭﺱ( ﺍﻧﻈﺮ ﻣﻠﺤﻖ ﺍﻹﺟﺎﺑﺎﺕ‬:‫ )ﺇﺭﺷﺎﺩ‬3.9 ‫( ﺍﻟﻨﻈﺮﻳﺔ‬12 ‫( ﺍﻧﻈﺮ ﻣﻠﺤﻖ ﺍﻹﺟﺎﺑﺎﺕ‬10 :‫ ﺍﻛﺘﺐ ﺑﺮﻫﺎﻧﹰﺎ ﻟﻜ ﱟﻞ ﻣﻤﺎ ﻳﺄﺗﻲ‬ 187  3-5 3.7 ‫( ﺍﻟﻨﻈﺮﻳﺔ‬10 ‫ ﺗﻮﺟﺪ ﺣﺎﻟﺘﺎﻥ ﳑﻜﻨﺘﺎﻥ( ﺍﻧﻈﺮ ﺍﻟﻬﺎﻣﺶ‬:‫ )ﺇﺭﺷﺎﺩ‬3.8 ‫( ﺍﻟﻨﻈﺮﻳﺔ‬11 A D ‫ ﺍﻧﻈﺮ ﻣﻠﺤﻖ ﺍﻹﺟﺎﺑﺎﺕ‬.13 ‫ﺍﺳﺘﻌﻤﻞ ﺍﻟﺸﻜﻞ ﺍﻟﻤﺠﺎﻭﺭ ﻟﻺﺟﺎﺑﺔ ﻋﻦ ﺍﻟﺴﺆﺍﻝ‬ E AB ⊥ BC, DC ⊥ BC  (13 AC BD BC AB DC  187  3-5 2   ABC , EFD  1  (11 ABC  .‫ﻗﺎﺋﻤﺎ ﺍﻟﺰﺍﻭﻳﺔ‬ ‫∠ ﻗﺎﺋﻤﺘﺎﻥ‬A, ∠E ‫ﻓﻴﻬﻤﺎ‬ ،‫ﻗﺎﺋﻤﺎ ﺍﻟﺰﺍﻭﻳﺔ‬ DEF ، ___ ___ ___ ___ AC DF , ∠C ∠F ،‫∠ ﻗﺎﺋﻤﺘﺎﻥ‬A , ∠D ‫ﻓﻴﻬﻤﺎ‬ CB DF , ∠B ∠F ABC DEF  ABC EFD  CF CD A BD E A BE F ‫ ﻗﺎﺋﻤﺎ‬ABC , DEF ‫ ﻧﻌﻠﻢ ﺃﻥ‬ ‫ ﻗﺎﺋﻤﺎ‬ABC , EFD_‫_ ﻧ_ﻌﻠﻢ_ﺃ_ﻥ‬_ ___ ___ ،AC DF ‫ ﻭ‬،‫∠ ﻗﺎﺋﻤﺘﺎﻥ‬A, ∠D ‫ ﻓﻴﻬﻤﺎ‬،‫ﺍﻟﺰﺍﻭﻳﺔ‬ . ∠B ∠F ، CB DF ،‫ﺍﻟﺰﺍﻭﻳﺔ‬ ‫∠؛ ﻷﻥ ﺟﻤﻴﻊ‬A ∠E ‫∠ ﻗﺎﺋﻤﺘﺎﻥ؛ ﻟﺬﺍ‬A, ∠E ‫∠ ؛ ﻷﻥ ﺟﻤﻴﻊ ﺍﻟﺰﻭﺍﻳﺎ‬A ∠D ‫ ﻭ‬.∠C ∠F ‫ﻭ‬ ABC EFD ‫ ﺇﺫﻥ‬.‫ﺍﻟﺰﻭﺍﻳﺎ ﺍﻟﻘﻮﺍﺋﻢ ﻣﺘﻄﺎﺑﻘﺔ‬ ‫ ﺑﺤﺴﺐ‬ABC DEF ‫ﺍﻟﻘﻮﺍﺋﻢ ﻣﺘﻄﺎﺑﻘﺔ؛ ﺇﺫﻥ‬ AAS. ‫ﺑﺤﺴﺐ‬ .ASA

  3 -5              (27)  (26)         3-5  3-5 ASA, AAS ASA, AAS :(AAS)        (ASA)        . (AAS) ‫ ﺯﺍﻭﻳﺔ – ﺯﺍﻭﻳﺔ – ﺿﻠﻊ‬:‫ﻳﻤﻜﻨﻚ ﺇﺛﺒﺎﺕ ﺗﻄﺎﺑﻖ ﻣﺜﻠﺜﻴﻦ ﻣﺴﺘﻌﻤ ﹰﻼ ﻧﻈﺮﻳﺔ‬ . (ASA) ‫ ﺯﺍﻭﻳﺔ – ﺿﻠﻊ – ﺯﺍﻭﻳﺔ‬:‫ﻳﻤﻜﻨﻚ ﺇﺛﺒﺎﺕ ﺗﻄﺎﺑﻖ ﻣﺜﻠﺜﻴﻦ ﻣﺴﺘﻌﻤ ﹰﻼ ﺍﻟﻤﺴ ﹼﻠﻤﺔ‬ ‫ﺇﺫﺍ ﻃﺎﺑﻘﺖ ﺯﺍﻭﻳﺘﺎﻥ ﻭﺿﻠﻊ ﻏﲑ ﳏﺼﻮﺭ ﺑﻴﻨﻬﲈ ﰲ ﻣﺜﻠﺚ ﻧﻈﺎﺋﺮﻫﺎ ﰲ ﻣﺜﻠﺚ ﺁﺧﺮ ﻓﺈﻥ ﺍﳌﺜﻠﺜﲔ‬ ‫ﺍﻟﺘﻄﺎﺑﻖ ﺑﺰﺍﻭﻳﺘﻴﻦ‬ .‫ ﻓﺈﻥ ﺍﳌﺜﻠﺜﲔ ﻣﺘﻄﺎﺑﻘﺎﻥ‬،‫ﺇﺫﺍ ﻃﺎﺑﻘﺖ ﺯﺍﻭﻳﺘﺎﻥ ﻭﺍﻟﻀﻠﻊ ﺍﳌﺤﺼﻮﺭ ﺑﻴﻨﻬﲈ ﰲ ﻣﺜﻠﺚ ﻧﻈﺎﺋﺮﳘﺎ ﰲ ﻣﺜﻠ ﹴﺚ ﺁﺧﺮ‬ ‫ﻣﺴ ﹼﻠﻤﺔ ﺍﻟﺘﻄﺎﺑﻖ ﺑﺰﺍﻭﻳﺘﻴﻦ‬ .‫( ﻣﺘﻄﺎﺑﻘﺎﻥ‬AAS) ‫ﻭﺿﻠﻊ ﻏﻴﺮ ﻣﺤﺼﻮ ﹴﺭ ﺑﻴﻨﻬﻤﺎ‬ (ASA) ‫ﻭﺿﻠﻊ ﻣﺤﺼﻮﺭ ﺑﻴﻨﻬﻤﺎ‬ :‫ﻟﻘﺪ ﺃﺻﺒﺢ ﻟﺪﻳﻚ ﺍﻵﻥ ﲬﺲ ﻃﺮﺍﺋﻖ ﻹﺛﺒﺎﺕ ﺗﻄﺎﺑﻖ ﻣﺜﻠﺜﲔ ﻫﻲ‬  SSS ‫• ﺍﳌﺴ ﹼﻠﻤﺔ‬ ASA ‫• ﺍﳌﺴ ﹼﻠﻤﺔ‬ AB .‫ﺍﻛﺘﺐ ﺑﺮﻫﺎ ﹰﻧﺎ ﺫﺍ ﻋﻤﻮﺩﻳﻦ‬ SAS ‫• ﺍﳌﺴ ﹼﻠﻤﺔ‬ AAS ‫• ﺍﳌﺴ ﹼﻠﻤﺔ‬ AB CD  B ‫ ﻣﺎ ﺍﻷﺿﻼﻉ ﺍﻟﻤﺘﻄﺎﺑﻘﺔ ﻓﻲ ﺍﻟﺸﻜﻞ؟‬،∠BCA ∠DCA :‫ﻓﻲ ﺍﻟﺸﻜﻞ ﺍﻟﻤﺠﺎﻭﺭ‬  ∆ABD ∆CDB ‫ﺇﺛﺒﺎﺕ ﺃﻥ‬   A 1 C ‫ ﺣﺘﻰ ﻳﻜﻮﻥ ﺍﻟﻤﺜﻠﺜﺎﻥ ﻣﺘﻄﺎﺑﻘﻴﻦ‬،‫ﻣﺎ ﺍﻟﻌﻨﺼﺮﺍﻥ ﺍﻟﻤﺘﻨﺎﻇﺮﺍﻥ ﺍﻵﺧﺮﺍﻥ ﺍﻟﻠﺬﺍﻥ ﻳﺘﻌ ﹼﻴﻦ ﺃﻥ ﻳﻜﻮﻧﺎ ﻣﺘﻄﺎﺑﻘﻴﻦ‬ DC 2 ‫؟‬AAS ‫ﻭﻓﻖ ﺍﻟﻨﻈﺮﻳﺔ‬ D .‫ ﻭﻓﻖ ﺧﺎﺻﻴﺔ ﺍﻻﻧﻌﻜﺎﺱ ﻟﻠﺘﻄﺎﺑﻖ‬AC AC       ‫ ﻳﻜﻮﻥ ﺍﻟﻀﻠﻊ ﺍﳌﺤﺼﻮﺭ ﺑﻴﻨﻬﲈ ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ؛ ﻟﺬﺍ ﻳﺘﻌﲔ ﺃﻥ ﺗﻜﻮﻥ‬AC ‫∠ ؛ ﻷﻥ‬2‫∠ ﻭ‬1 ‫ﻻ ﻳﻤﻜﻦ ﺃﻥ ﻳﻜﻮﻥ ﺍﻟﻌﻨﴫﺍﻥ ﺍﻵﺧﺮﺍﻥ ﳘﺎ‬ ‫( ﻣﻌﻄﻴﺎﺕ‬1 AB CD (1 .AAS ‫∆ ﻭﻓﻖ ﺍﻟﻨﻈﺮﻳﺔ‬ABC ∆ADC ‫ ﻭﻋﻨﺪﻫﺎ ﻳﻜﻮﻥ‬،∠B ∠D ‫( ﻧﻈﺮﻳﺔ ﺍﻟﺰﺍﻭﻳﺘﲔ ﺍﳌﺘﺒﺎﺩﻟﺘﲔ ﺩﺍﺧﻠ ﹼﹰﻴﺎ‬2 ∠CBD ∠ADB (2  ‫( ﻧﻈﺮﻳﺔ ﺍﻟﺰﺍﻭﻳﺘﲔ ﺍﳌﺘﺒﺎﺩﻟﺘﲔ ﺩﺍﺧﻠ ﹰﹼﻴﺎ‬3 ∠ABD ∠BDC (3 A :‫ﺍﻛﺘﺐ ﺑﺮﻫﺎ ﹰﻧﺎ ﻣﻦ ﺍﻟﻨﻮﻉ ﺍﻟﻤﺤ ﹼﺪﺩ ﻓﻲ ﻛﻞ ﻣﻦ ﺍﻟﺴﺆﺍﻟﻴﻦ ﺍﻵﺗﻴﻴﻦ‬  ‫( ﺧﺎﺻﻴﺔ ﺍﻻﻧﻌﻜﺎﺱ ﻟﻠﺘﻄﺎﺑﻖ‬4 BD BD (4 .‫( ﺑﺮﻫﺎﻥ ﺫﻭ ﻋﻤﻮﺩﻳﻦ‬1 B C BC EF   ASA ‫( ﺍﳌﺴ ﹼﻠﻤﺔ‬5 ∆ABD ∆CDB (5 D AB ED .∆ABC ∠C ∠F  ∆DEF ‫ﺇﺛﺒﺎﺕ ﺃﻥ‬  :‫ﺍﻛﺘﺐ ﺑﺮﻫﺎ ﹰﻧﺎ ﻣﻦ ﺍﻟﻨﻮﻉ ﺍﳌﺤ ﹼﺪﺩ ﰲ ﻛ ﱟﻞ ﻣﻦ ﺍﻟﺴﺆﺍﻟﲔ ﺍﻵﺗﻴﲔ‬ : EF  BD .‫( ﺑﺮﻫﺎﻥ ﺣﺮ‬2 V .‫( ﺑﺮﻫﺎﻥ ﺫﻭ ﻋﻤﻮﺩﻳﻦ‬1 RT   AE ‫ ﺗﻨﺼﻒ‬CD  ∠S ∠V  U .SV ‫ ﻧﻘﻄﺔ ﻣﻨﺘﺼﻒ‬T  (1 BC EF , AB ED (1 A C E ∠E ∠BCA  AB CD S ∠C ∠F    (2 ∠ABC ∠DEF (2 ∆ABC ∆CDE ‫ﺇﺛﺒﺎﺕ ﺃﻥ‬  Geo-SG04-0∆5-R0T7S-86∆01U81TV ‫ﺇﺛﺒﺎﺕ ﺃﻥ‬  AAS   (3 ∆ABC ∆DEF (3 : Geo-SG04-05-08-860181 : S .‫( ﺑﺮﻫﺎﻥ ﺗﺴﻠﺴﻠﻲ‬2   AE  CD ∠E ∠BCA :     R T ∠STU ‫ ﺗﻨ ﹼﺼﻒ‬TR ;∠S ∠U   AC CD       AC CE U  (1  ∠S ∠V (1      ∠A ∠DCE  SV   T ∠SRT ∠URT ‫ﺇﺛﺒﺎﺕ ﺃﻥ‬  ∠STR ∠UTR ∠STU ‫ ﺗﻨﺼﻒ‬TR .ASA    ∆ABC ∆CDE     (2 ST TV (2 ‫ﺗﻌﺮﻳﻒ ﻣﻨﺼﻒ ﺍﻟﺰﺍﻭﻳﺔ‬ ‫ﻣﻌﻄﻴﺎﺕ‬      (3 ∠RTS ∠VTU (3 ∠SRT ∠URT SRT URT ∠S ∠U ASA   (4 ∆RTS ∆UTV (4 ‫ﺍﻟﻌﻨﺎﴏ ﺍﳌﺘﻨﺎﻇﺮﺓ ﰲ ﺍﳌﺜﻠﺜﲔ‬ AAS ‫ﻣﻌﻄﻴﺎﺕ‬ ‫ﺍﳌﺘﻄﺎﺑﻘﲔ ﺗﻜﻮﻥ ﻣﺘﻄﺎﺑﻘﺔ‬ 27 RT RT ‫ﺧﺎﺻﻴﺔ ﺍﻻﻧﻌﻜﺎﺱ ﻟﻠﺘﻄﺎﺑﻖ‬ 3   3 26    (29)   ( 2 8 )        3-5   3-5 ASA, AAS ASA, AAS ‫ ﺃﻧﻮﺍﻉ ﻣﺨﺘﻠﻔﺔ ﻣﻦ ﺍﻟﻘﻄﻊ‬5 ‫ﺍﺳﺘﻌﻤﻞ ﻣﻬﻨﺪﺱ ﺩﻳﻜﻮﺭ‬ (4 ،‫( ﺃﺭﺍﺩ ﺃﺧﻮﺍﻥ ﺍﻗﺘﺴﺎﻡ ﻗﻄﻌﺔ ﺍﻷﺭﺽ ﺍﻟﺘﺎﻟﻴﺔ ﺑﺎﻟﺘﺴﺎﻭﻱ‬1 J .‫( ﺍﻛﺘﺐ ﺑﺮﻫﺎ ﹰﻧﺎ ﺗﺴﻠﺴﻠ ﹰﹼﻴﺎ‬1 ￯‫ ﺇﺫﺍ ﻭﺟﺪ ﺇﺣﺪ‬،‫ﺍﻟﺰﺟﺎﺟﻴﺔ ﺍﻟﻤﺜﻠﺜﺔ ﺍﻟﺸﻜﻞ ﻟﻌﻤﻞ ﺩﻳﻜﻮﺭ ﻟﻠﺤﺎﺋﻂ‬ A ،‫ﻗﻄﻊ ﺍﻟﺰﺟﺎﺝ ﻣﻜﺴﻮﺭﺓ ﻭﺍﻟﺒﺎﻗﻲ ﻣﻨﻬﺎ ﻫﻮ ﺍﻟﺠﺰﺀ ﺍﻟﻤﺒﻴﻦ ﺃﺩﻧﺎﻩ‬ B K L ∠N ∠L D N . JK MK C .∆JKN M ‫ﺃﻥ‬ ‫ﺇﺛﺒﺎﺕ‬  ∆MKL ‫ ﻳﻘﺴﻢ ﺍﻷﺭﺽ ﺇﱃ‬DB ‫ﻓﻘﺎﻡ ﺃﺣﺪﳘﺎ ﺑﻮﺿﻊ ﺣ ﱟﺪ ﻓﺎﺻ ﹴﻞ‬ ‫ﻓﻬﻞ ﻳﺴﺘﻄﻴﻊ ﻣﻌﺮﻓﺔ ﻫﺬﻩ ﺍﻟﻘﻄﻌﺔ ﻣﻦ ﺃﻱ ﻧﻮﻉ ﻣﻦ ﺍﻷﻧﻮﺍﻉ‬ ‫ ﲬﻦ ﺍﳌﻌﻠﻮﻣﺔ ﺍﻟﺘﻲ ﻛﺎﻧﺖ ﻟﺪﻳﻪ ﺣﺘﻰ ﻳﺘﻴﻘﻦ‬،‫ﻗﻄﻌﺘﲔ ﻣﺘﻄﺎﺑﻘﺘﲔ‬ : .‫ﺍﳋﻤﺴﺔ ﺍﳌﺨﺘﻠﻔﺔ ﺍﻟﺘﻲ ﺍﺳﺘﻌﻤﻠﻬﺎ؟ ﺑ ﱢﺮﺭ ﺇﺟﺎﺑﺘﻚ‬ ‫ﻣﻦ ﺃﻥ ﻗﻄﻌﺘﻲ ﺍﻷﺭﺽ ﻣﺘﻄﺎﺑﻘﺘﺎﻥ؟‬ JKN MKL ∠N ∠L ASA  ∠D  ∠B     DB :  AAS ‫ﻣﻌﻄﻴﺎﺕ‬              .ASA  .      JK MK ‫ﻣﻌﻄﻴﺎﺕ‬ ∠JKN ∠MKL ‫ﺍﻟﺰﺍﻭﻳﺘﺎﻥ ﺍﳌﺘﻘﺎﺑﻠﺘﺎﻥ ﺑﺎﻟﺮﺃﺱ ﻣﺘﻄﺎﺑﻘﺘﺎﻥ‬ ‫ﺗﺤﺘﺎﺝ ﺇﺩﺍﺭﺓ ﺣﺪﻳﻘﺔ ﺇﻟﻰ ﻏﻄﺎﺀ ﻋﻠﻰ ﻫﻴﺌﺔ‬ (5 ،￯‫ﺃﺭﺍﺩ ﺧﺎﻟﺪ ﺻﻨﻊ ﺍﻟﻬﺮﻡ ﺍﻟﻤﺒﻴﻦ ﺑﺎﻟﺸﻜﻞ ﻣﻦ ﺍﻟﻮﺭﻕ ﺍﻟﻤﻘﻮ‬ (2 B ‫( ﺍﻛﺘﺐ ﺑﺮﻫﺎ ﹰﻧﺎ ﺗﺴﻠﺴﻠ ﹼﹰﻴﺎ‬2 ‫ ﻃﻮﻝ‬،‫ﻣﺜﻠﺚ ﻟﺘﻐﻄﻴﺔ ﺣﻘﻞ ﻋﻠﻰ ﻫﻴﺌﺔ ﻣﺜﻠﺚ ﻣﺘﻄﺎﺑﻖ ﺍﻷﺿﻼﻉ‬ D   AB C B .‫ ﻣﺜﻠﺜﺎ ﹴﺕ ﻣﺘﻄﺎﺑﻘﺔ ﻋﲆ ﺍﻷﻗﻞ‬3 ‫ﻓﺎﺳﺘﻨﺘﺞ ﺃﻥ ﻋﻠﻴﻪ ﺍﻟﻘﻴﺎﻡ ﺑﻘ ﱢﺺ‬ A .200 ft ‫ﺿﻠﻌﻪ‬ .‫ﻓﻬﻞ ﺍﺳﺘﻨﺘﺎﺟﻪ ﺻﺤﻴﺢ؟ ﺑ ﱢﺮﺭ ﺇﺟﺎﺑﺘﻚ‬ .∠A ∠C AD CD ‫ ﻗﻴﺎﺱ ﻛ ﱟﻞ‬،‫( ﺇﺫﺍ ﹸﻭﺟﺪ ﻏﻄﺎﺀ ﻋﻠﻰ ﻫﻴﺌﺔ ﻣﺜﻠﺚ ﻓﻴﻪ ﺯﺍﻭﻳﺘﺎﻥ‬a            ‫ﺍﻟﻌﻨﺎﴏ ﺍﳌﺘﻨﺎﻇﺮﺓ‬ .∠ABC ‫ ﺗﻨ ﹼﺼﻒ‬DB ‫ ﻓﻬﻞ ﻫﺬﺍ‬،200 ft ‫ ﻭﻃﻮﻝ ﺃﺣﺪ ﺃﺿﻼﻋﻪ‬، 60° ‫ﻣﻨﻬﲈ‬ .ASA   ‫ﰲ ﺍﳌﺜﻠﺜﲔ ﺍﳌﺘﻄﺎﺑﻘﲔ‬ C .AD CD ‫ﺇﺛﺒﺎﺕ ﺃﻥ‬  .‫ﺍﻟﻐﻄﺎﺀ ﻣﻨﺎﺳﺐ ﻟﺬﻟﻚ ﺍﳊﻘﻞ؟ ﻭ ﱢﺿﺢ ﺇﺟﺎﺑﺘﻚ‬ ‫ﻟﺪ￯ ﻧﺠﺎﺭ ﻗﻄﻌﺔ ﺧﺸﺒﻴﺔ ﻋﻠﻰ ﺷﻜﻞ ﻣﺜﻠﺚ ﻣﺘﻄﺎﺑﻖ‬ (3 ‫ﺗﻜﻮﻥ ﻣﺘﻄﺎﺑﻘﺔ‬ :  ‫ ﻓﺮﺳﻢ‬،‫ ﺇﺫﺍ ﺃﺭﺍﺩ ﺃﻥ ﻳﻘﺴﻤﻬﺎ ﺇﻟﻰ ﻗﺴﻤﻴﻦ ﻣﺘﻄﺎﺑﻘﻴﻦ‬،‫ﺍﻷﺿﻼﻉ‬   60°         ‫ ﻭﻋﻤﻮﺩﻳ ﹰﺔ ﻋﻠﻰ ﺍﻟﻀﻠﻊ‬A ‫ﺑﺎﻟﻘﻠﻢ ﻗﻄﻌ ﹰﺔ ﻣﺴﺘﻘﻴﻤ ﹰﺔ ﻣﻦ ﺍﻟﻨﻘﻄﺔ‬ ABD CBD AB CB  ASA   60°    ‫ ﻓﻬﻞ ﻣﺎ ﻗﺎﻡ ﺑﻪ ﺍﻟﻨﺠﺎﺭ ﻳﻀﻤﻦ ﺗﻄﺎﺑﻖ ﺍﻟﻤﺜﻠﺜﻴﻦ‬، A ‫ﺍﻟﻤﻘﺎﺑﻞ ﻟﹺـ‬ ASA ‫ﻣﻌﻄﻴﺎﺕ‬     AAS .‫ﺍﻟﻨﺎﺗﺠﻴﻦ؟ ﺑ ﱢﺮﺭ ﺫﻟﻚ‬ ∠A ∠C ‫ﻣﻌﻄﻴﺎﺕ‬ ‫( ﺇﺫﺍ ﻋﻠﻤﺖ ﺃﻥ ﻗﻴﺎﺱ ﻛ ﱟﻞ ﻣﻦ ﺍﻟﺰﻭﺍﻳﺎ ﺍﻟﺜﻼﺙ ﰲ ﻏﻄﺎ ﹴﺀ ﻣﺜﻠﺚ‬b A ‫ ﻓﻬﻞ ﻳﻜﻮﻥ ﻫﺬﺍ ﺍﻟﻐﻄﺎﺀ ﻣﻨﺎﺳ ﹰﺒﺎ ﻟﺬﻟﻚ‬،60° ‫ﺍﻟﺸﻜﻞ ﻳﺴﺎﻭﻱ‬ ∠ABD ∠CBD ∠ABC ‫ ﺗﻨﺼﻒ‬DB ‫ﺗﻌﺮﻳﻒ ﻣﻨﺼﻒ ﺍﻟﺰﺍﻭﻳﺔ‬ ‫ﻣﻌﻄﻴﺎﺕ‬ ‫ﺍﳊﻘﻞ ﺑﺎﻟﴬﻭﺭﺓ؟‬ D E .‫( ﺍﻛﺘﺐ ﺑﺮﻫﺎ ﹰﻧﺎ ﺣ ﹼﹰﺮﺍ‬3           G F DE FG ∠E ∠G BDC ∆DFG ∆FDE ‫ﺇﺛﺒﺎﺕ ﺃﻥ‬          :             ∠EDF ∠GFD DE F G   :             ∆DFG ∆FDE       DF FD  ∠E ∠G           .AAS  .   3 28  29  3  3 187A