In the given figure, XY || BC. Given that AX = 3 cm, XB = 1.5 cm and BC = 6 cm.

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

(ii) Calculate XY.

(i) By basic proportionality theorem,

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

Since, XY || BC

$\therefore \dfrac{AY}{YC} = \dfrac{AX}{XB} \\[1em] = \dfrac{3}{1.5} \\[1em] = \dfrac{2}{1}.$

Hence, $\dfrac{AY}{YC} = \dfrac{2}{1}$.

(ii) In ΔAXY and ΔABC,

∠AXY = ∠ABC [Corresponding angles are equal]

∠XAY = ∠BAC [Common ]

∴ ΔAXY ∼ ΔABC.

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

$\Rightarrow \dfrac{AX}{AB} = \dfrac{XY}{BC} \\[1em] \Rightarrow \dfrac{AX}{AX + XB} = \dfrac{XY}{BC} \\[1em] \Rightarrow \dfrac{3}{3 + 1.5} = \dfrac{XY}{6} \\[1em] \Rightarrow \dfrac{3}{4.5} = \dfrac{XY}{6} \\[1em] \Rightarrow XY = \dfrac{3}{4.5} \times 6 \\[1em] \Rightarrow XY = 4 \text{ cm.}$

Hence, XY = 4 cm.