Giải thích các bước giải:
a.
\(\begin{array}{l}
{S_{ABD}} + {S_{BDC}} = {S_{ABCD}} = 30\\
\frac{1}{2}.{h_B}.AD + \frac{1}{2}.{h_C}.BC = 30\\
\frac{1}{2}.{h_B}.AD + \frac{1}{2}.{h_B}.\frac{{AD}}{2} = 30\\
\frac{1}{2}.{h_B}.AD = 20\\
{S_{ABD}} = 20c{m^2}
\end{array}\)
\(\begin{array}{l}
{S_{AMD}} + {S_{BMD}} = {S_{ABD}} = 20\\
\frac{1}{2}.{h_D}.AM + \frac{1}{2}.{h_D}.BM = 20\\
\frac{1}{2}.{h_D}.AM + \frac{1}{2}.{h_D}.\frac{{3AM}}{2} = 20\\
\frac{1}{2}.{h_D}.AM = 8\\
{S_{AMD}} = 8c{m^2}
\end{array}\)
b.
\(\begin{array}{l}
{S_{AMND}} + {S_{BCNM}} = {S_{ABCD}} = 30\\
{S_{AMND}} = 3{S_{BCNM}}\\
\to {S_{AMND}} = \frac{{45}}{2}c{m^2},{S_{BCNM}} = \frac{{15}}{2}c{m^2}\\
{S_{AMND}} = {S_{AMD}} + {S_{MND}} = \frac{{45}}{2}\\
\to {S_{MND}} = \frac{{45}}{2} - 8 = \frac{{29}}{2}c{m^2}
\end{array}\)
\(\begin{array}{l}
\begin{array}{*{20}{l}}
{{S_{ABC}} + {S_{ADC}} = {S_{ABCD}} = 30}\\
{\frac{1}{2}.{h_A}.BC + \frac{1}{2}.{h_D}.AD = 30}\\
{\frac{1}{2}.{h_A}.BC + \frac{1}{2}.{h_A}.2.BC = 30}\\
{\frac{1}{2}.{h_A}.BC = 10}\\
{{S_{ABC}} = 10c{m^2}}
\end{array}\\
{\rm{\;}}\begin{array}{*{20}{l}}
{{S_{AMC}} + {S_{BMC}} = {S_{ABC}} = 10}\\
{\frac{1}{2}.{h_C}.AM + \frac{1}{2}.{h_C}.BM = 10}\\
{\frac{1}{2}.{h_C}.\frac{2}{3}.BM + \frac{1}{2}.{h_C}.BM = 10}\\
{\frac{1}{2}.{h_C}.BM = 6}\\
{{S_{BMC}} = 6c{m^2}}
\end{array}
\end{array}\)
\(\begin{array}{l}
{S_{BCNM}} = {S_{BCM}} + {S_{MNC}} = \frac{{15}}{2}\\
\to {S_{MNC}} = \frac{{15}}{2} - 6 = \frac{3}{2}c{m^2}
\end{array}\)
\(\begin{array}{l}
\frac{{{S_{MNC}}}}{{{S_{MND}}}} = \frac{{\frac{1}{2}.{d_M}.CN}}{{\frac{1}{2}.{d_M}.DN}} = \frac{{\frac{{29}}{2}}}{{\frac{3}{2}}} = \frac{{29}}{3}\\
\to \frac{{CN}}{{DN}} = \frac{{29}}{3}
\end{array}\)
