Ratio of perimeters of similar triangles

In $\Delta BCD$ and $\Delta BAC$, $\angle B$ is common and $\angle BCD = \angle BAC$. Therefore, $\Delta BCD \sim \Delta BAC$. Ratio of sides: $\frac{BC}{BA} = \frac{DB}{BC} = \frac{CD}{AC}$. $\frac{12}{AB} = \frac{9}{12} \implies AB = 16$ cm. $AD = AB - DB = 16 - 9 = 7$ cm. $\frac{6}{AC} = \frac{9}{12} \implies AC = 8$ cm. Perimeter of $\Delta ADC = AD + DC + AC = 7 + 6 + 8 = 21$ cm. Perimeter of $\Delta BDC = BD + DC + BC = 9 + 6 + 12 = 27$ cm. Ratio $= \frac{21}{27} = \frac{7}{9}$.