Question

In the figure of $\Delta ABC,DE\parallel AB$.

If $AD=2x,DC=x+3,BE=2x-1$ and $CE=x$, then find the value of $x$.

Answer 1

Applying basic proportionality theorem, we get

$\frac{DC}{AD}=\frac{EC}{BE}\frac{x+3}{2x}=\frac{x}{2x-1}(x+3)(2x-1)=x\left(2x\right)2x^2-x+6x-3=2x^{2}5x=3x=\frac{3}{5}$