Fie CE⊥AB
In ΔCEB dr in E, avem ∡B=60⇒ m(∡BCE)=30°⇒ 2EB=BC
EB=6:2=3cm
CD=AE=3cm⇒ AB=AE+EB=3+3=6cm
In ΔCEB aplicam Pitagora
BC²=CE²+EB²
36=CE²+9
CE=3√3 cm
In ΔAEC aplicam Pitagora
AC²=CE²+AE²
AC²=27+9=36
AC=6cm
In ΔADC dr in D, DM mediana ⇒ DM=AC:2=6:2=3cm