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