Question

[Vũ Minh Tú]

Answer 1

[Akai Haruma]

Lời giải:

$\frac{S_{DNC}}{S_{DBC}}=\frac{NC}{BC}=\frac{NC}{NC+BN}=\frac{NC}{NC+NC}=\frac{1}{2}$

$\Rightarrow S_{DNC}=\frac{1}{2}S_{DBC}=\frac{1}{2}.\frac{1}{2}S_{ABCD}=\frac{1}{4}S_{ABCD}(1)$

$\frac{S_{AMD}}{S_{ABD}}=\frac{AM}{AB}=\frac{AM}{AM+MB}=\frac{AM}{AM+3AM}=\frac{1}{4}$

$\Rightarrow S_{AMD}=\frac{1}{4}S_{ABD}=\frac{1}{4}.\frac{1}{2}S_{ABCD}=\frac{1}{8}S_{ABCD}(2)$

Từ $(1); (2)\Rightarrow S_{AMD}+S_{DNC}=(\frac{1}{8}+\frac{1}{4})S_{ABCD}=\frac{3}{8}S_{ABCD}$
$\Rightarrow S_{MBND}=S_{ABCD}-(S_{AMD}+S_{DNC})=S_{ABCD}-\frac{3}{8}S_{ABCD}=\frac{5}{8}S_{ABCD}$