Macedonian Mathematical Competitions

Because $\angle BDC = \angle ABC = \beta$ and $\angle DCB = \angle BCA = \gamma$ we have that $\triangle BDC \sim \triangle ABC$. From the similarity we have $\overline{BC}:\overline{DC} = \overline{AC}:\overline{BC}$, from where we get $\overline{BC}^2 = \overline{DC} \cdot \overline{AC} = 9 \cdot (7+9) = 144$. Hence $\overline{BC} = 12\text{cm}$.

From $\triangle BDC \sim \triangle ABC$ we also have $\overline{BD}:\overline{DC} = \overline{AB}:\overline{BC}$ or $\overline{BD}:\overline{AB} = \overline{DC}:\overline{BC} = 9:12 = 3:4$. Now we obtain $\overline{BD}:\overline{BA} = 3:4$.