In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.

(i) Prove that ΔACD is similar to ΔBCA.

(ii) Find BC and CD.

(iii) Find the area of ΔACD : area of ΔABC.

(i) In ∆ACD and ∆BCA,

∠DAC = ∠ABC [Given]

∠ACD = ∠BCA [Common angles]

∴ ∆ACD ~ ∆BCA [By AA]

Hence, proved that ∆ACD ~ ∆BCA.

(ii) Since, ∆ACD ~ ∆BCA

We know that,

Corresponding sides of similar triangle are proportional.

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

Also,

$\Rightarrow \dfrac{CD}{AC} = \dfrac{AD}{AB} \\[1em] \Rightarrow \dfrac{CD}{4} = \dfrac{5}{8} \\[1em] \Rightarrow CD = \dfrac{4 \times 5}{8} \\[1em] \Rightarrow CD = \dfrac{5}{2} = 2.5 \text{ cm}.$

Hence, BC = 6.4 cm and CD = 2.5 cm.

(iii) As, ∆ACD ~ ∆BCA

We know that,

The areas of two similar triangles are proportional to the squares of their corresponding sides.

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

Hence, area of ∆ACD : area of ∆ABC = 25 : 64.