In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.

Find:

(i) $\dfrac{AY}{YC}$

(ii) $\dfrac{YC}{AC}$

(iii) XY

(i) By basic proportionality theorem,

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

$\therefore \dfrac{AY}{YC} = \dfrac{AX}{XB} \\[1em] = \dfrac{9}{4.5} \\[1em] = \dfrac{2}{1}.$

Hence, $\dfrac{AY}{YC} = 2.$

(ii) Given,

$\Rightarrow \dfrac{AY}{YC} = \dfrac{2}{1}$

Let AY = 2x and YC = x.

From figure,

AC = AY + YC = 2x + x = 3x.

$\therefore \dfrac{YC}{AC} = \dfrac{x}{3x} = \dfrac{1}{3}$.

Hence, $\dfrac{YC}{AC} = \dfrac{1}{3}$.

(iii) In △AXY and △ABC,

∠AXY = ∠ABC [Corresponding angles are equal]

∠A = ∠A [Common]

∴ △AXY ~ △ABC.

Since, corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{AX}{AB} = \dfrac{XY}{BC} \\[1em] \Rightarrow \dfrac{AX}{AX + XB} = \dfrac{XY}{BC} \\[1em] \Rightarrow \dfrac{9}{9 + 4.5} = \dfrac{XY}{18} \\[1em] \Rightarrow \dfrac{9}{13.5} = \dfrac{XY}{18} \\[1em] \Rightarrow XY = \dfrac{9}{13.5} \times 18 \\[1em] \Rightarrow XY = 12 \text{ cm}. \\[1em]$

Hence, XY = 12 cm.