Two circles with centers $B$ and $D$ have radii $DA=8\text{cm}$ and $BC=x\text{cm}$, respectively. $AC$ is tangent to both circles. If $DB$ and $AC$ intersect at the point $E$, $AE=12\text{cm}$ and $EC=18\text{cm}$, then find the value of $x$ (in $\text{cm}$).

Since $AC$ is a common tangent to both circles at points $A$ and $C$ respectively, the radius from the center to the point of tangency is perpendicular to the tangent line.
Therefore, $\angle DAE=90^{\circ }$ and $\angle BCE=90^{\circ }$.

In $△DAE$ and $△BCE$:

1. $\angle DAE=\angle BCE=90^{\circ }$
2. $\angle AED=\angle CEB$ (Vertically opposite angles)

By AA (Angle-Angle) similarity criterion:

$$△DAE\sim △BCE$$

Since the triangles are similar, the ratios of their corresponding sides are equal:

$$\frac{DA}{BC}=\frac{AE}{EC}$$

Substitute the given values into the equation:

$$\frac{8}{x}=\frac{12}{18}$$

Simplify the fraction on the right side:

$$\frac{8}{x}=\frac{2}{3}$$

Cross-multiplying to solve for $x$:

$$2x=8\times 3$$

$$2x=24$$

$$x=12\text{cm}$$

Therefore, the value of $x$ is $12$.