Page 143 - Engineering

‫ﺍﳍﻨﺪﺳﺔ )ﺍﳉﺰء ﺍﻷﻭﻝ(‬ ‫‪١٣٠‬‬

‫‪ .CD = DE = EB ،BC = 3AC‬ﻣﺎ ﻗﻴﺎس اﻟﺰاوﻳﺔ ‪ FDC‬؟‬

‫‪A‬‬

‫‪F‬‬

‫‪B‬‬ ‫‪EXD‬‬ ‫‪C‬‬

‫)د( ‪150°‬‬ ‫)ج( ‪135°‬‬ ‫)ب( ‪125°‬‬ ‫)أ( ‪115°‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )ج(‪ :‬ارﺳﻢ ‪ FX‬ﻋﻤﻮدﻳﺎً ﻋﻠﻰ ‪ .BC‬اﻵن‪،‬‬

‫‪.(AA) △FXB ∼ △ACB‬‬

‫ﲟﺎ أن ‪ BF = 1‬ﻓﺈن ‪ . BX = FX = 1‬ﻟﻨﻔﺮض أن ‪ .AC = a‬ﻋﻨﺪﺋﺬ‪،‬‬
‫‪BC AC 2‬‬ ‫‪BA 2‬‬

‫‪ .BX = 3a ،BC = 3a ،FX = 1a‬ﻣﻦ ذﻟﻚ ﻧﺮى أن‪:‬‬

‫‪22‬‬

‫‪EX = BX − BE = 3 a − a = a = FX‬‬
‫‪22‬‬

‫‪XD = BD − BX = a = FX‬‬
‫‪2‬‬

‫إذن‪ FX ،‬ﻣﺘﻮﺳﻂ ﰲ ‪ △DEF‬وﻃﻮﻟﻪ ﻳﺴﺎوي ‪ . 1 ED‬إذن‪ .EFD = 90° ،‬وﲟﺎ‬

‫‪2‬‬

‫أن ‪ (SAS) △FXD ≡ △FXE‬ﻓﺈن ‪ . FD = FE‬و‪‬ﺬا ﻳﻜﻮن ‪△DEF‬‬

‫ﻣﺘﺴﺎوي اﻟﺴﺎﻗﲔ‪ .‬إذن‪ FDE = 45° ،‬وﻳﻜﻮن ‪.FDC = 180° − 45° = 135°‬‬

‫)‪ [AHSME 1966] (٧١‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ‪ .MD PC ،AM = MB ،‬ﻣﺎ‬
‫ﻗﻴﻤﺔ ]‪ [BPD‬؟‬

‫] ‪[ABC‬‬