In the figure (2) given below, AF, BE and CD are parallel lines. Given that AF = 7.5 cm, CD = 4.5 cm, ED = 3 cm and BE = x and AE = y. Find the values of x and y.

Consider △AEF and △CED

∠AEF = ∠CED [Vertically opposite angles]

∠F = ∠C [Alternate angles are equal]

So, by AA rule of similarity △AEF ~ △CED

Then,

$\dfrac{AF}{CD} = \dfrac{AE}{ED} \\[1em] \dfrac{7.5}{4.5} = \dfrac{y}{3} \\[1em] y = \dfrac{22.5}{4.5} \\[1em] y = 5.$

Consider △ABE and △ACD

∠A = ∠A [Common angles]

∠ABE = ∠ACD [Alternate angles are equal]

So, by AA rule of similarity △ABE ~ △ACD.

Then,

$\dfrac{EB}{CD} = \dfrac{AE}{AD} \\[1em] \dfrac{x}{4.5} = \dfrac{y}{AE + ED} \\[1em] \dfrac{x}{4.5} = \dfrac{y}{y + 3} \\[1em] x = \dfrac{4.5 \times 5}{8} \\[1em] x = \dfrac{22.5}{8} \\[1em] x = \dfrac{225}{80} = \dfrac{45}{16} \\[1em] x = 2\dfrac{13}{16}.$

Hence, the value of x = $2\dfrac{13}{16}$ cm and y = 5 cm.