In the given figure, DE || BC.

(i) If AD = 3.6 cm, AB = 9 cm and AE = 2.4 cm, find EC.

(ii) If $\dfrac{AD}{DB} = \dfrac{3}{5}$ and AC = 5.6 cm, find AE.

(iii) If AD = x cm, DB = (x − 2) cm, AE = (x + 2) cm and EC = (x − 1) cm, find the value of x.

By basic proportionality theorem,

A line drawn parallel to a side of triangle divides the other two sides proportionally.

(i) Given,

AD = 3.6 cm

AB = 9 cm

AE = 2.4 cm

Since, DE || BC by basic proportionality theorem,

$\Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC} \\[1em] \Rightarrow \dfrac{AD}{AB - AD} = \dfrac{AE}{EC} \\[1em] \Rightarrow \dfrac{3.6}{9 - 3.6} = \dfrac{2.4}{EC}\\[1em] \Rightarrow \dfrac{3.6}{5.4} = \dfrac{2.4}{EC}\\[1em] \Rightarrow EC = \dfrac{5.4 \times 2.4}{3.6}\\[1em] \Rightarrow EC = 3.6 \text{ cm.}$

Hence, EC = 3.6 cm.

(ii) Given,

$\dfrac{AD}{DB} = \dfrac{3}{5}$

AC = 5.6 cm

Since, DE || BC by basic proportionality theorem,

$\Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC} \\[1em] \Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{AC - AE} \\[1em] \Rightarrow \dfrac{3}{5} = \dfrac{AE}{5.6 - AE}\\[1em] \Rightarrow 3 \times (5.6 - AE) = 5AE \\[1em] \Rightarrow 16.8 - 3AE = 5AE \\[1em] \Rightarrow 5AE + 3AE = 16.8 \\[1em] \Rightarrow 8AE = 16.8 \\[1em] \Rightarrow AE = \dfrac{16.8}{8} \\[1em] \Rightarrow AE = 2.1 \text{ cm.}$

Hence, AE = 2.1 cm.

(iii) Given,

AD = x cm

DB = (x − 2) cm

AE = (x + 2) cm

EC = (x − 1) cm

Since, DE || BC by basic proportionality theorem,

$\Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC} \\[1em] \Rightarrow \dfrac{x}{(x - 2)} = \dfrac{(x + 2)}{(x - 1)} \\[1em] \Rightarrow x(x - 1) = (x + 2) (x - 2) \\[1em] \Rightarrow x^2 - x = (x^2 - (2)^2) \\[1em] \Rightarrow x^2 - x = x^2 - 4 \\[1em] \Rightarrow x^2 - x - x^2 + 4 = 0 \\[1em] \Rightarrow - x + 4 = 0 \\[1em] \Rightarrow x = 4 \text{ cm.}$

Hence, x = 4 cm.