Page 133 - Engineering

‫ﺍﳍﻨﺪﺳﺔ )ﺍﳉﺰء ﺍﻷﻭﻝ(‬ ‫‪١٢٠‬‬

‫)ب( ) ‪AD = (AB )(BD‬‬ ‫)أ( ‪AD = AE‬‬

‫) ‪EC (BE )(BC‬‬ ‫‪EC DC‬‬

‫)د( ) ‪AD = (AE )(BD‬‬ ‫)ج( ‪AD = AB‬‬

‫) ‪EC (DC )(BE‬‬ ‫‪EC BC‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )ب(‪:‬‬

‫‪A‬‬

‫‪D‬‬
‫‪E‬‬

‫‪BC‬‬

‫‪ BD‬ﻳﻨﺼﻒ اﻟﺰاوﻳﺔ ‪ .ABE‬إذن‪. AD = AB ،‬‬

‫‪DE BE‬‬

‫‪ BE‬ﻳﻨﺼﻒ اﻟﺰاوﻳﺔ ‪ .DBC‬إذن‪. DE = DB ،‬‬

‫‪EC BC‬‬

‫‪. AD‬‬ ‫=‬ ‫) ‪DE (AB/BE‬‬ ‫)‪(AB)(BD‬‬ ‫ذﻟﻚ ﻳﻜﻮن‪،‬‬ ‫وﻣﻦ‬
‫=‬
‫) ‪EC DE(BC/BD) (BE)(BC‬‬

‫)‪ [AHSME 1952] (٥٩‬رﲰﻨﺎ ﻋﻠﻰ اﻟﻮﺗﺮ ‪ AB‬ﻟﻠﻤﺜﻠﺚ اﻟﻘﺎﺋﻢ اﻟﺰاوﻳﺔ ‪△ABC‬‬

‫ﻣﺜﻠﺜﺎً آﺧﺮ ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ ‪ △ABD‬وﺗﺮﻩ ‪ .AB‬إذا ﻛﺎن ‪،AC = b ،BC = 1‬‬
‫‪ AD = 2‬ﻓﻤﺎ ﻃﻮل ‪ BD‬؟‬

‫)ج( ‪) b2 + 1‬د( ‪b2 + 3‬‬ ‫)ب( ‪b2 − 1‬‬ ‫)أ( ‪b2 − 3‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )أ(‪:‬‬