如图,已知△ABC∽△A1B1C1,相似比为K（K＞1）且△ABC的三边长分别为a,b,c(a>b>c).△A1B1C1

(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
再问： 第二题解答过程是什么意思