In the figure, given below, AB, CD and EF are parallel lines. Given AB = 7.5 cm, DC = y cm, EF = 4.5 cm, BC = x cm and CE = 3 cm, calculate the values of x and y.

In ∆BEF,

DC || FE

So, by basic proportionality theorem,

$\dfrac{BD}{DF} = \dfrac{BC}{CE} = \dfrac{x}{3}$

$\dfrac{BD}{DF} = \dfrac{x}{3}$ …..(1)

Let BD = ax and DF = 3a.

From figure,

BF = BD + DF = ax + 3a = a(x + 3).

$\dfrac{BF}{DF} = \dfrac{a(x + 3)}{3a} = \dfrac{x + 3}{3}$ …..(2)

In ∆AFB and CDF,

∠AFB = ∠CFD [Common angles]

∠ABF = ∠CDF [Corresponding angles are equal]

∴ ∆AFB ~ ∆CDF [By AA]

Since, corresponding sides of similar triangles are proportional.

$\Rightarrow \dfrac{BF}{DF} = \dfrac{AB}{CD} \\[1em] \Rightarrow \dfrac{BF}{DF} = \dfrac{7.5}{y} …..(3) \\[1em] \text{From (2) and (3) we get}, \\[1em] \Rightarrow \dfrac{x + 3}{3} = \dfrac{7.5}{y} \\[1em] \Rightarrow y = \dfrac{22.5}{x + 3} \space …..(4)$

In ∆BCD and ∆BEF ,

∠FBE = ∠DBC [Common angles]

∠CDB = ∠EFB [Corresponding angles are equal]

∴ ∆BEF ~ ∆BCD [By AA]

Since, corresponding sides of similar triangles are proportional.

$\dfrac{BD}{BF} = \dfrac{CD}{FE} = \dfrac{y}{4.5}$

$\dfrac{BD}{BF} = \dfrac{y}{4.5}$

Let BD = ay and BF = 4.5a

From figure,

DF = BF - BD = 4.5a - ay = a(4.5 - y).

$\therefore \dfrac{BD}{DF} = \dfrac{ay}{a(4.5 - y)} = \dfrac{y}{4.5 - y}$ …..(5)

From (1) and (5) we get,

$\Rightarrow \dfrac{x}{3} = \dfrac{y}{4.5 - y}$

Substituting value of y from (4) in above equation we get,

$\Rightarrow \dfrac{x}{3} = \dfrac{\dfrac{22.5}{x + 3}}{4.5 - \dfrac{22.5}{x + 3}} \\[1em] \Rightarrow \dfrac{x}{3} = \dfrac{\dfrac{22.5}{x + 3}}{\dfrac{4.5x + 13.5 - 22.5}{x + 3}} \\[1em] \Rightarrow \dfrac{x}{3} = \dfrac{22.5}{4.5x - 9} \\[1em] \Rightarrow \dfrac{x}{3} = \dfrac{22.5}{3(1.5x - 3)} \\[1em] \Rightarrow x = \dfrac{22.5}{1.5x - 3} \\[1em] \Rightarrow x = \dfrac{22.5}{1.5(x - 2)} \\[1em] \Rightarrow x = \dfrac{15}{x - 2} \\[1em] \Rightarrow x(x - 2) = 15 \\[1em] \Rightarrow x^2 - 2x - 15 = 0 \\[1em] \Rightarrow x^2 - 5x + 3x - 15 = 0 \\[1em] \Rightarrow x(x - 5) + 3(x - 5) = 0 \\[1em] \Rightarrow (x + 3)(x - 5) = 0 \\[1em] \Rightarrow x = - 3 \text{ or } x = 5.$

Since, side of triangle cannot be negative. So, x = 5 cm.

Substituting value of x in (4) we get,

$\Rightarrow y = \dfrac{22.5}{x + 3} \\[1em] \Rightarrow y = \dfrac{22.5}{5 + 3} \\[1em] \Rightarrow y = \dfrac{22.5}{8} = 2.8125 \text{ cm}$

Hence, x = 5 cm and y = 2.8125 cm