\(\frac{S_{ABD}}{S_{ADC}}=\frac{BD}{CD}=\frac{AB}{AC}=\frac{c}{b}\Rightarrow\frac{S_{ABD}}{S_{ADC}}+1=\frac{c}{b}+1\)
\(\Rightarrow\frac{S_{ABC}}{S_{ADC}}=\frac{b+c}{b}\Rightarrow S_{ADC}=\frac{S_{ABC}.b}{b+c}\)(1)
Lại có: AM là trung tuyến nên \(S_{AMC}=\frac{1}{2}S_{ABC}\)(2)
Lấy (1) trừ (2) được \(S_{ADC}-S_{AMC}=S_{ABC}.\left(\frac{b}{b+c}+\frac{1}{2}\right)\)
\(\frac{\Rightarrow S_{ADM}}{S_{ABC}}=\frac{3b+c}{2\left(b+c\right)}\).