Ukrainian Mathematical Competitions

It is clear that $\angle AMD=20^{\circ }$, $\angle DBM=150^{\circ }$. Let's construct equilateral triangle $△KMD$, then points $M$, $B$, $D$ are on the circle centered at $K$ (fig. 8.76). After that $KM=KB$, $\angle KMB=80^{\circ }$, from where $\angle MBK=80^{\circ }$, but where $\angle MBC=80^{\circ }$ as well, which implies that points $B$, $C$, $K$ are on the same line.

Then $△AMD=△BKM$, as isosceles with equal sides and angles at the base. Therefore $MB=AD$. From similarity of $△MBC\sim △MAD$ we have that $\frac{BC}{AD}=\frac{MB}{MA}$ $\Rightarrow$ $AD\cdot BM=BC\cdot MA$, hence we obtain that

Fig. 45

$$A{D}^2=BC(MB+AB)=BC(AD+AB).$$