Giải thích các bước giải:
Ta có :
$\begin{split}S_{CDQN}&=S_{BCD}-S_{QBM}\\&=\dfrac 12-\dfrac{BM}{BC}S_{QBC}\\&=\dfrac 12-\dfrac 12S_{QBC}\\&=\dfrac 12-\dfrac 12.\dfrac{BQ}{BD}S_{BCD}\\&=\dfrac 12-\dfrac 12.\dfrac{BM}{BM+AD}S_{BCD}\\&=\dfrac 12-\dfrac 12.\dfrac{1}{3}.\dfrac 12=\dfrac{5}{12}\end{split}$