Page 144 - Engineering

‫ﺍﳌﺜﻠﺜﺎﺕ ‪١٣١‬‬

‫‪C‬‬
‫‪D‬‬

‫‪A PM‬‬ ‫‪B‬‬

‫‪1‬‬ ‫)د(‬ ‫‪1‬‬ ‫)ج(‬ ‫‪1‬‬ ‫)ب(‬ ‫‪1‬‬ ‫)أ(‬
‫‪6‬‬ ‫‪4‬‬ ‫‪3‬‬ ‫‪2‬‬

‫‪ MD‬ﻓﺈن ] ‪) [MDC ] = [MDP‬ﺑﻌﺪ رﺳﻢ‬ ‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )أ(‪ :‬ﲟﺎ أن ‪PC‬‬

‫اﳌﺴﺘﻘﻴﻢ ‪ .(CM‬اﻵن‪،‬‬

‫] ‪[BPD] = [BMD] + [MDP ] = [BMD] + [MDC ] = [BMC ] = 1[ABC‬‬
‫‪2‬‬

‫ﻷن ‪ CM‬ﻣﺘﻮﺳﻂ‪ .‬إذن‪. [BPD] = 1 ،‬‬

‫‪[ABC ] 2‬‬

‫)‪ [AHSME 1966] (٧٢‬ﰲ اﳌﺜﻠﺚ ‪ △ABC‬اﳌﺮﻓﻖ‪ AM ،‬و ‪ CN‬ﻣﺘﻮﺳﻄﺎن‬
‫ﻳﺘﻘﺎﻃﻌﺎن ﰲ اﻟﻨﻘﻄﺔ ‪ Q ،AP = PC .O‬ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ‪ MP‬ﻣﻊ ‪.CN‬‬

‫إذا ﻛﺎن ‪ [OMQ] = n‬ﻓﺈن ] ‪ [ABC‬ﺗﺴﺎوي‪:‬‬

‫‪C‬‬

‫‪P QM‬‬
‫‪O‬‬

‫‪AN B‬‬

‫)د( ‪24n‬‬ ‫)ج( ‪21n‬‬ ‫)ب( ‪18n‬‬ ‫)أ( ‪16n‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )د(‪ :‬ﻗﺎﻋﺪة اﳌﺜﻠﺚ ‪ △OMQ‬ﺗﺴﺎوي‬