Cho tứ giác \(ABCD\) có \(AB = 4\;\,{\rm{cm,}}\;\,AD = 6\;\,{\rm{cm,}}\;\,BD = 8\;\,{\rm{cm,}}\;\,BC = 12\;\,{\rm{cm,}}\;\,CD = 16\;\,{\rm{cm}}{\rm{.}}\)

a) Đúng.

Ta có: \(MN\parallel BC\) nên theo định lí Thalès, ta có: \(\frac{{AM}}{{AB}} = \frac{{AN}}{{AC}} = \frac{{MN}}{{BC}}\).

b) Sai.

Ta có: \(\frac{{AM}}{{AB}} = \frac{{AN}}{{AC}} = \frac{{MN}}{{BC}}\) (cmt) nên \(\Delta ABC \sim \Delta AMN\) (c.c.c).

c) Đúng.

Từ a) ta có: \(\frac{{AM}}{{AB}} = \frac{{AN}}{{AC}} = \frac{{MN}}{{BC}}\) hay \(\frac{{AM}}{4} = \frac{{AN}}{6} = \frac{{MN}}{8} = \frac{{BM}}{6} = \frac{{AM + BM}}{{4 + 6}} = \frac{{AB}}{{10}} = \frac{4}{{10}}\).

Do đó, \(AN = \frac{4}{{10}}AC = \frac{4}{{10}}.6 = 2,4{\rm{ cm}}\).

            \(MN = \frac{4}{{10}}.8 = 3,2{\rm{ cm}}\).

d) Đúng.

Ta có \(\Delta AMN \sim \Delta ABC\) theo tỉ số đồng dạng \(k = \frac{4}{{10}} = \frac{2}{5}\) (từ câu b).

Do đó, \(\frac{{{S_{ANM}}}}{{{S_{ABC}}}} = \frac{{M{N^2}}}{{B{C^2}}} = \frac{{{2^2}}}{{{5^2}}} = \frac{4}{{25}}\).

Đáp án: 16

Vì \(ME\parallel AB\) nên \(\Delta EMC \sim \Delta ABC\) nên \(\frac{{MC}}{{BC}} = \frac{{EM}}{{AB}} = \frac{{EC}}{{AC}} = \frac{2}{3}\).

Do đó, \({P_{EMC}} = \frac{2}{3}{P_{ABC}} = \frac{2}{3} \cdot 24 = 16\,\,\left( {{\rm{cm}}} \right)\)