Giải thích các bước giải:
Ta có :
$\dfrac{NB}{ND}=\dfrac{BM}{AD}=\dfrac{1}{2}\to \dfrac{NB}{BD}=\dfrac{1}{3}$
$S_{MNDC}=S_{BDC}-S_{NMB}=S_{BDC}-\dfrac{1}{2}S_{NCB}=S_{BDC}-\dfrac{1}{2}.\dfrac{1}{3}S_{BCD}=\dfrac{5}{6}S_{DBC}=\dfrac{5}{12}S_{ABCD}=\dfrac{5}{12}S$