Ta có \(\displaystyle {{AM} \over {AC}} = {{10} \over {15}} = {2 \over 3}\)
\(\displaystyle {{AN} \over {AB}} = {8 \over {12}} = {2 \over 3}\)
\( \Rightarrow \displaystyle {{AM} \over {AC}} = {{AN} \over {AB}}= {2 \over 3}\)
Xét \(∆ AMN\) và \(∆ ACB\) có:
+) \(\widehat A\) chung
+) \(\displaystyle {{AM} \over {AC}} = {{AN} \over {AB}}= {2 \over 3}\)
\( \Rightarrow ∆ AMN\) đồng dạng \(∆ ACB\) (c.g.c)
\( \Rightarrow \displaystyle {{AN} \over {AB}} = {{MN} \over {BC}}\)
\( \Rightarrow \displaystyle MN = {{AN.BC} \over {AB}} = {{8.18} \over {12}} = 12\)\(\; (cm).\)
