Estonian Mathematical Olympiad

Opposite angles $AMB$ and $DMC$ equal. Also notice that $\angle ABM = \angle ABD = \angle ACD = \angle MCD$, as $ABD$ and $ACD$ are subtended to the same arc. Therefore the triangles $AMB$ and $DMC$ are similar. Hence $\frac{BA}{BM} = \frac{CD}{CM}$, from which $BM = BA \cdot \frac{CM}{CD}$. Analogously the opposite angles $AMD$ and $BMC$ are equal. By the property of inscribed angles $\angle ADM = \angle ADB = \angle ACB = \angle MCB$. Hence, the triangles $AMD$ and $BMC$ are similar and $\frac{AD}{AM} = \frac{BC}{BM}$. In summary, $AD = \frac{AM \cdot BC}{BM} = \frac{AM \cdot 5 \text{ mm}}{2 \text{ mm}} \cdot 0.6 = 6 \text{ mm}$.

Fig. 1