An angle bisector of a triangle divides the opposite side of the triangle into segments 6 cm and 8 cm long. A second side of the triangle measures 7 cm long. Find the longest and shortest possible lengths of the third side of the triangle. Round answers to the nearest tenth. (1 point)
Let the triangle be $ABC$, where $BC=7$. Let the angle bisector of angle $A$ intersect side $BC$ at point $D$. By using angle bisector theorem, we can conclude that
\[\frac{AB}{BD} = \frac{AC}{CD}.\]
Since $BD=\frac{6}{8}\cdot BC = \frac{6}{8}\cdot 7 = \frac{21}{4}$ and $CD=\frac{8}{8}\cdot BC = \frac{8}{8}\cdot 7 = 7$, we can substitute these into the angle bisector theorem to get
\[\frac{AB}{\frac{21}{4}} = \frac{AC}{7}.\]
Hence, we have
\[AB = \frac{21}{4}\cdot\frac{AC}{7} = \frac{3}{4}\cdot AC.\]
By using the triangle inequality, we also know that $AC+AB>BC$, or equivalently, $\frac{7}{4}AC>\frac{21}{4}$, or $AC>3$. Now, we find the two possible lengths of $AC$: we set $\frac{3}{4}\cdot AC = \frac{21}{4}+7$ to find the maximum value of $AC$, and we set $\frac{3}{4}\cdot AC = \frac{21}{4}-7$ to find the minimum value of $AC$. We obtain
\[AC_{\text{max}} = \frac{21}{3} = 7 \text{\quad and\quad } AC_{\text{min}}= \frac{21}{3}-7 = 0.\]
Hence, the longest and shortest possible lengths of side $AC$ are $\boxed{7\text{ cm}}$ and $\boxed{0\text{ cm}}$, respectively.