In $\Delta ABC$, $AD$ is the internal bisector of $\angle A$, which meets side $BC$ at $D$. If $BD=5\text{cm}$ and $BC=7.5\text{cm}$, then what is the ratio $AB:AC$?

According to the Internal Angle Bisector Theorem, the internal bisector of an angle of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.
Therefore, in $\Delta ABC$:

$$\frac{AB}{AC}=\frac{BD}{DC}$$

Given:

* $BD=5\text{cm}$
* $BC=7.5\text{cm}$

We can find the length of segment $DC$:

$$DC=BC-BD$$

$$DC=7.5\text{cm}-5\text{cm}=2.5\text{cm}$$

Now substitute the values of $BD$ and $DC$ into the ratio:

$$\frac{AB}{AC}=\frac{5}{2.5}=\frac{50}{25}=\frac{2}{1}$$

Thus, the ratio $AB:AC=2:1$.