In the figure (ii) given below, ∠ABC = ∠DAC and AB = 8 cm, AC = 4 cm, AD = 5 cm.

(i) Prove that △ACD is similar to △BCA.

(ii) Find BC and CD.

(iii) Find area of △ACD : area of △ABC.

(i) Considering △ACD and △BCA.

∠C = ∠C (Common angles)

∠ABC = ∠DAC (Given)

Hence, by AA axiom △ACD ~ △BCA.

(ii) Since triangles are similar, hence the ratio of corresponding sides will be equal

$\Rightarrow \dfrac{AC}{BC} = \dfrac{AD}{AB} \\[1em] \Rightarrow \dfrac{4}{BC} = \dfrac{5}{8} \\[1em] \Rightarrow BC = \dfrac{4 \times 8}{5} \\[1em] \Rightarrow BC = \dfrac{32}{5} \\[1em] \Rightarrow BC = 6.4 \text{ cm}.$

Similarly,

$\Rightarrow \dfrac{CD}{CA} = \dfrac{AD}{AB} \\[1em] \Rightarrow \dfrac{CD}{4} = \dfrac{5}{8} \\[1em] \Rightarrow CD = \dfrac{4 \times 5}{8} \\[1em] \Rightarrow CD = \dfrac{20}{8} \\[1em] \Rightarrow CD = 2.5 \text{ cm}.$

Hence, the length of BC = 6.4 cm and CD = 2.5 cm.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ACD}}{\text{Area of △ABC}} = \dfrac{AD^2}{AB^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ADC}}{\text{Area of △ABC}} = \dfrac{5^2}{8^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ADC}}{\text{Area of △ABC}} = \dfrac{25}{64}.$

Hence, the ratio of area of △ACD : area of △ABC = 25 : 64.