Question:
In the given figure, DE ∥ BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.
Solution:
$\because D E \| B C$
$\therefore \frac{A D}{D B}=\frac{A E}{E C} \quad$ (Basic proportionality theorem)
$\frac{x}{3 x+4}=\frac{x+3}{3 x+19}$
$\Rightarrow x(3 x+19)=(x+3)(3 x+4)$
$\Rightarrow 3 x^{2}+19 x=3 x^{2}+4 x+9 x+12$
$\Rightarrow 19 x-13 x=12$
$\Rightarrow 6 x=12$
$\Rightarrow x=2$