What value of $x$ will make $D E ∥ A B$ , in the given figure?
$A D = 8 x + 9, C D = x + 3, B E = 3 x + 4, C E = x .$

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Solution

In $△ A B C, D E ∥ B C$
$C D / D A = C E / B E$ [Thales theorem ]
$\begin{matrix} (x + 3) / (8 x + 9) = x / (3 x + 4) \Rightarrow (x + 3) (3 x + 4) = x (8 x + 9) \Rightarrow 3 x^{2} + 4 x + 9 x + 12 = 8 x^{2} + 9 x \Rightarrow 3 x^{2} + 13 x + 12 = 8 x^{2} + 9 x \Rightarrow 5 x^{2} - 4 x - 12 = 0 \Rightarrow 5 x^{2} - 10 x + 6 x - 12 = 0 \Rightarrow 5 x (x - 2) + 6 (x - 2) = 0 \Rightarrow (x - 2) (5 x + 6) = 0 x - 2 = 0 or 5 x + 6 = 0 \end{matrix}$
Therefore, $x = 2$ or $x = - 6 / 5$ .

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In the adjoining figure, DE || AB,

AD = 7 cm, CD = 5 cm and BC = 18 cm

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