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 ?

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 ? Correct Answer 7 : 9

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