According to question,

AB = AC,         ∠B = ∠C
∠A + ∠B + ∠C = 180°
∠A + 2∠B = 180°
∠A = 180° - 70°
∠A = 110°
Note : In isosceles triangle median bisects the opposite side and make angle 90° on opposite side. It also bisects the vertex angle.
∠BAD = $$\frac{{\angle A}}{2}$$
∠BAD = $$\frac{{{{110}^ \circ }}}{2}$$
∠BAD = 55°

∵ AB = AC
   Point D is the incenter
∴ ∠BDC = 90° + $$\frac{1}{2}$$ ∠A
              = 90° + $$\frac{1}{2}$$ × 80°
              = 90° + 40°
              = 130°

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, $$\eqalign{ & \frac{{{\text{Perimeter of Delta ADC}}}}{{{\text{Perimeter of Delta BDC}}}} \cr & = \frac{{6 + 7 + 8}}{{9 + 6 + 12}} \cr & = \frac{{21}}{{27}} \cr & = \frac{7}{9} \cr} $$