Equilateral Triangle Length of AD

$BD + CD = 3 \implies BD + 2BD = 3 \implies BD = 1\text{ cm}$. Let E be the midpoint of BC (foot of altitude AE). $BE = 1.5\text{ cm} \implies DE = BE - BD = 1.5 - 1 = 0.5\text{ cm}$. Altitude $AE = \frac{\sqrt{3}}{2} \times 3 = \frac{3\sqrt{3}}{2}\text{ cm}$. In right $\triangle ADE$, $AD^2 = AE^2 + DE^2 = \frac{27}{4} + \frac{1}{4} = \frac{28}{4} = 7 \implies AD = \sqrt{7}\text{ cm}$.