a: \(BD=\dfrac{2\cdot12\cdot6}{12+6}\cdot cos60=4\left(cm\right)\)
c: Xét ΔABC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)
=>12^2+6^2-AC^2=2*12*6*cos120
=>AC=6 căn 7(cm)
\(AM^2=\dfrac{2\left(252+36\right)-12^2}{4}=108\)
hay \(AM=6\sqrt{3}\left(cm\right)\)