Ta có:
$\dfrac{AM}{AB} = \dfrac{AN}{AC} = \dfrac{1}{3}$
$\Rightarrow MN//BC$
$\Rightarrow d(M;BC) = d(N;BC)$ (với $d$ là khoảng cách)
$\Rightarrow \dfrac{d(M;BC).BC}{2} = \dfrac{d(N;BC).BC}{2}$
$\Rightarrow S_{∆BMC} = S_{∆BNC}$
$\Rightarrow S_{∆BMC} - S_{∆BOC}= S_{∆BNC} - S_{∆BOC}$
$\Rightarrow S_{∆BMO} = S_{∆CNO}$
$\Rightarrow S_{∆BMO} + S_{∆MON} = S_{∆CNO} + S_{∆MON}$
$\Rightarrow S_{∆BMN} = S_{∆CMN}$
$\Rightarrow S_{∆BMN} + S_{∆AMN} = S_{∆CMN} + S_{∆AMN}$
$\Rightarrow S_{∆BAN} = S_{∆CAM}$
Xét $∆AMN$ và $∆ABC$ có:
$\dfrac{AM}{AB} = \dfrac{AN}{AC} = \dfrac{1}{3}$ $(gt)$
$\widehat{BAC}:$ góc chung
Do đó $∆AMN\sim ∆ABC$ $(c.g.c)$
$\Rightarrow \dfrac{S_{∆AMN}}{S_{∆ABC}} = (\dfrac{AM}{AN})^{2} = \dfrac{1}{9}$
$\Rightarrow S_{MNCB} = \dfrac{8S_{∆ABC}}{9} = \dfrac{8.36}{9} = 32 cm^{2}$