In the figure given below, AB ∥ DE, AC = 3cm, CE = 7.5cm and BD = 14cm. Calculate CB and DC.

In the given figure,

AB ∥ DE, AC = 3cm, CE = 7.5cm, BD = 14cm.

From the figure,

∠ACB = ∠DCE [Vertically opposite angles]
∠BAC = ∠CED [Alternate angles]

Then, by AA rule of similarity, △ABC ~ △CDE.

So,

$\Rightarrow \dfrac{AC}{CE} = \dfrac{BC}{CD} \\[1em] \Rightarrow \dfrac{3}{7.5} = \dfrac{BC}{CD} \\[1em] \Rightarrow 7.5BC = 3CD \qquad \text{[….Eq 1]} \\[1em]$

∵ BD = 14 cm, Let BC = x cm

∴ From fig, CD = (14 - x) cm.

Putting these values of BC and CD in equation 1 we get,

$\Rightarrow 7.5x = 3(14 - x) \\[1em] \Rightarrow 7.5x = 42 - 3x \\[1em] \Rightarrow 7.5x + 3x = 42 \\[1em] \Rightarrow 10.5x = 42 \\[1em] \Rightarrow x = \dfrac{42}{10.5} \\[1em] \Rightarrow x = 4.$

∴ x = 4 and 14 - x = 10.

Hence, CB = 4cm and DC = 10cm.