Từ B kẻ đường cao $BH$
Ta có:
$S_{ABC} = \dfrac{1}{2}AC.BH= 200 \, cm^2$
$S_{ABN} = \dfrac{1}{2}AN.BH = \dfrac{1}{2}.\dfrac{2}{3}AC.BH$
$= \dfrac{2}{3}.\left(\dfrac{1}{2}AC.BH\right) = \dfrac{2}{3}S_{ABC} = \dfrac{2}{3}.200 = \dfrac{400}{3} \, cm^2$
$S_{BNC} = S_{ABC} - S_{ABN} = \dfrac{1}{3}S_{ABC} = \dfrac{200}{3} \, cm^2$