In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, find

(i) BC

(ii) DC

(iii) area of △ACD : area of △BCA

(i) Considering △ABC and △ACD,

∠C = ∠C (Common angles)

∠ABC = ∠CAD (Given)

Hence by AA axiom △ABC ~ △ACD. Since triangles are similar hence the ratio of the corresponding sides will be equal

$\therefore \dfrac{AB}{AD} = \dfrac{BC}{AC} \\[1em] \Rightarrow \dfrac{5}{4} = \dfrac{BC}{3} \\[1em] \Rightarrow BC = \dfrac{15}{4} \\[1em] \Rightarrow BC = 3.75$

Hence, the length of BC = 3.75 cm.

(ii) Since triangles △ABC and △ACD are similar hence the ratio of the corresponding sides will be equal.

$\therefore \dfrac{AB}{AD} = \dfrac{AC}{DC} \\[1em] \Rightarrow \dfrac{5}{4} = \dfrac{3}{DC} \\[1em] \Rightarrow DC = \dfrac{12}{5} \\[1em] \Rightarrow DC = 2.4$

Hence, the length of DC = 2.4 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.

$\therefore \dfrac{\text{Area of △ACD}}{\text{Area of △BCA}} = \dfrac{AD^2}{AB^2} \\[1em] = \dfrac{4^2}{5^2} \\[1em] = \dfrac{16}{25}$

Hence, the ratio of area of △ACD : area of △BCA is 16 : 25.