-△ABC vuông tại A, AM là trung tuyến \(\Rightarrow BM=AM=CM=\dfrac{1}{2}BC\)
\(\Rightarrow\)△ABM cân tại M mà \(\widehat{BAM}=60^0\)\(\Rightarrow\)△ABM đều.
\(\Rightarrow AB=BM=\dfrac{1}{2}BC\Rightarrow BC=6\left(cm\right)\)
-△ABC vuông tại A \(\Rightarrow AB^2+AC^2=BC^2\Rightarrow AC=\sqrt{BC^2-AB^2}=\sqrt{6^2-3^2}=\sqrt{27}\left(cm\right)\)
\(\Rightarrow S_{ABC}=\dfrac{1}{2}.AB.AC=\dfrac{1}{2}.3.\sqrt{27}=\dfrac{1}{2}.3.3.\sqrt{3}=\dfrac{9\sqrt{3}}{2}\left(cm^2\right)\)
M là trung điểm BC\(\Rightarrow S_{ACM}=\dfrac{1}{2}.S_{ABC}=\dfrac{1}{2}.\dfrac{9\sqrt{3}}{2}=\dfrac{9\sqrt{3}}{4}\left(cm^2\right)\)
-△ABC có BD phân giác \(\Rightarrow\dfrac{AD}{CD}=\dfrac{AB}{CB}\Rightarrow\dfrac{AD}{AB}=\dfrac{CD}{CB}=\dfrac{AD+CD}{AB+CB}=\dfrac{AC}{AB+CB}\Rightarrow\dfrac{CD}{AC}=\dfrac{CB}{AB+CB}\)\(\dfrac{S_{DMC}}{S_{AMC}}=\dfrac{CD}{AC}=\dfrac{CB}{AB+CB}=\dfrac{6}{3+6}=\dfrac{2}{3}\)
\(\Rightarrow\dfrac{S_{DMC}}{\dfrac{9\sqrt{3}}{4}}=\dfrac{2}{3}\)
\(\Rightarrow S_{DMC}=\dfrac{3\sqrt{3}}{2}\left(cm^2\right)\)
