Giải thích các bước giải:
Ta có $MC=2MB\to \dfrac{MB}{MC}=\dfrac12\to\dfrac{MB}{MB+MC}=\dfrac1{1+2}=\dfrac13$
$\to\dfrac{MB}{BC}=\dfrac13\to MB=\dfrac13BC$
Kẻ $AD//BC, D\in BN$
$\to \dfrac{AD}{BC}=\dfrac{AN}{CN}=2$
$\to AD=2BC$
Lại có $AD//CB \to \dfrac{AE}{EM}=\dfrac{AD}{BM}=\dfrac{2BC}{\dfrac13BC}=6$
$\to \dfrac{ME}{AE}=\dfrac16$
$\to\dfrac{ME}{ME+AE}=\dfrac1{6+1}$
$\to \dfrac{ME}{MA}=\dfrac17$
$\to ME=\dfrac17MA$
$\to S_{BME}=\dfrac17S_{ABM}=\dfrac17\cdot \dfrac{MB}{BC}S_{ABC}=\dfrac17\cdot \dfrac13S_{ABC}$
$\to S_{BME}=\dfrac1{21}S_{ABC}=\dfrac1{21}S$