In $\triangle ABC, DE || BC$. If $AE = (2x + 1)$ cm, $EC= 4$ cm, $AD= (x + 1)$ cm and $DB = 3$ cm, then value of $x$ is:

In $\angle ABC, DE || BC$. If $AE = (2x + 1)$ cm, $EC= 4$ cm, $AD= (x + 1)$ cm and $DB = 3$ cm, then value of $x$ is:

Given that $ \displaystyle DE \parallel BC $, by the **Basic Proportionality Theorem (Thales' Theorem)**, we have:

$ \displaystyle \frac{AD}{DB} = \frac{AE}{EC} $

Substituting the given values:

$ \displaystyle \frac{x + 1}{3} = \frac{2x + 1}{4} $

Cross multiplying:

$ \displaystyle (x + 1) \times 4 = (2x + 1) \times 3 $

$ \displaystyle 4x + 4 = 6x + 3 $

Rearranging:

$ \displaystyle 4 + 1 = 6x - 4x $

$ \displaystyle 5 = 2x $

$ \displaystyle x = \frac{5}{2} $

Thus, the value of $ \displaystyle x $ is **$ \displaystyle \frac{5}{2} $**.