问题描述: 已知△ABC相似△A1B1C1,相似比为k(k>1),且△ABC的三边长分别为a、b、c(a>b>c),△A1B1C1的三边长分别为a1、b1、c1 1个回答 分类:数学 2014-12-14 问题解答: 我来补答 (1)证:Q△ ABC ∽△ A1 B1C1 ,且相似比为 k ( k > 1),a = k,a = ka1.∴ a1 又Q c = a1,a = kc.取 a = 8,b = 6,c = 4,同时取a1 = 4,b1 = 3,c1 = 2.此时 a b c = = = 2,△ ABC ∽△ A1 B1C1 且 c = a1.∴ a1 b1 c1不存在这样的△ ABC 和△ A1 B1C1 .理由如下:若 k = 2,a = 2a1,b = 2b1,c = 2c1.则 又Q b = a1,= b1 ,c ∴ a = 2a1 = 2b = 4b1 = 4c,∴ b = 2c.∴ b + c = 2c + c < 4c = a ,而 b + c > a,-1- 故不存在这样的△ ABC 和△ A1 B1C1 ,使得 k = 2.回答完毕 展开全文阅读