Consider the triangle shown in the figure where BC = 12 cm, DB = 9 cm, CD = 6 cm and ∠BCD = ∠BAC. What is the ratio of the perimeter of the triangle ADC to that of the triangle BDC ?

Answer: Option A

Here, ∠ACB = c + 180 - (2c - b) = 180 - (b + c)
So, We can say that ΔBCD and ΔABC will be similar.
According to property of similarity,
$$\frac{\text{AB}}{12}=\frac{12}{9}$$
Hence,
AB = 16
$$\frac{\text{AC}}{6}=\frac{12}{9}$$
AC = 8
Hence, AD = 7 and AC = 8
Now,
$$\begin{array}{rl}& \frac{\text{Perimeter of Delta ADC}}{\text{Perimeter of Delta BDC}}\\ & =\frac{6+7+8}{9+6+12}\\ & =\frac{21}{27}\\ & =\frac{7}{9}\end{array}$$