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MATH CE PP 1991 #53

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問題： 圖片參考：http://i.minus.com/jrTexVXv4RZmc.jpg 解： 圖片參考：http://i.minus.com/jCA60hhFNUUKL.jpg 我想問解果個做法詳細係點？ 佢d比例點搵出黎？

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• bb好難掃風
• 做愛問題 , 呢JJ好硬@1@
• S$ 24.90 即係港幣幾多錢-@1@

 

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Draw a line AH perpendicular to DC and meet DC at H. Area of △ADK=DKxAH/2 [底x高/2] Area of △AKC=KCxAH/2 Area of △ADK/Area of △AKC =(DKxAH/2)/(KCxAH/2) =DK/KC AD=2DB AD:DB=2:1 Draw a line DE parallel to AM and meet BC at E. △BDE~△BAM ME:EB=AD:DB=2:1 EM:BM=2:3 M is the mid-point of BC. BM=MC EM:MC=2:3 △CKM~△CDE DK:KC=EM:MC=2:3 Area of △ADK/Area of △AKC=DK/KC=2/3 2012-03-13 17:56:12 補充： Method 2: M is the mid-point of BC. Area of △ABM= Area of △ACM Area of △KBM= Area of △KCM Area of △ABM-Area of △KBM =Area of △ACM-Area of △KCM Area of △AKB=Area of △AKC AD=2DB Area of △ADK:Area of △DBK=2:1 Area of △ADK:Area of △AKB=2:3 Area of △ADK:Area of △AKC=2:3

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