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In the given figure, $XZ$ is parallel to $BC$. $AZ = 3$ cm, $ZC = 2$ cm, $BM =3$ cm and $MC = 5$ cm. Find the length of $XY$.
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As $XZ \parallel BC$ Therefore $\frac{AX}{XB} = \frac{AZ}{ZC} = \frac{3}{2}$ (i)
$\triangle AXY \sim \triangle ABM$
$\Rightarrow \frac{AX}{AB} = \frac{XY}{BM}$ or $\frac{XY}{3} = \frac{3}{5}$
$\Rightarrow XY = \frac{9}{5}$ or $1.8$ cm
$\triangle AXY \sim \triangle ABM$
$\Rightarrow \frac{AX}{AB} = \frac{XY}{BM}$ or $\frac{XY}{3} = \frac{3}{5}$
$\Rightarrow XY = \frac{9}{5}$ or $1.8$ cm