Giải thích các bước giải:
a.Ta có $AD=2BC\to S_{ACD}=2S_{ABC}$
Mà $S_{ABCD}=S_{ACD}+S_{ABC}=30\to S_{ACD}=20, S_{ABC}=10$
Lại có $BC//AD\to S_{ABD}=S_{ACD}=20, S_{ABC}=S_{DBC}=10$
Mà $\dfrac{AM}{BM}=\dfrac23$
$\to \dfrac{AM}{AM+BM}=\dfrac{2}{2+3}\to \dfrac{AM}{AB}=\dfrac25$
$\to \dfrac{BM}{AB}=\dfrac{AB-AM}{AB}=1-\dfrac{AM}{AB}=\dfrac35$
$\to S_{CBM}=\dfrac35S_{CAB}=6, S_{DAM}=\dfrac25S_{ABD}=4$
b.Trường hợp $1: S_{BCNM}=3S_{MNDA}$
Mà $S_{ABCD}=S_{BCNM}+S_{MNDA}\to S_{BCNM}=\dfrac34\cdot S_{ABCD}=\dfrac{45}{2}$
$\to S_{MNDA}=\dfrac{15}{2}$
$\to S_{MCN}=S_{BCNM}-S{BMC}=\dfrac{33}{2}$
$S_{MND}=S_{MNDA}-S_{ADM}=\dfrac72$
Trường hợp $2: S_{MNDA}=3S_{BCNM}$
Tương tự câu $TH1\to S_{MNDA}=\dfrac{45}2, S_{BCNM}=\dfrac{15}2$
$\to S_{MCN}=S_{BCNM}-S{BMC}=\dfrac{3}{2}$
$S_{MND}=S_{MNDA}-S_{ADM}=\dfrac{27}2$
