In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$.
We can set $AE=ED=m$. Set $BD=k$, therefore $CD=3k, AC=4k$. Thereafter, by Stewart's Theorem on $\triangle ACD$ and cevian $CE$, we get $2m^2+14=25k^2$. Also apply Stewart's Theorem on $\triangle CEB$ with cevian $DE$. After simplification, $2m^2=17-6k^2$. Therefore, $k=1, m=\frac{\sqrt{22}}{2}$. Finally, note that (using [] for area) $[CED]=[CAE]=3[EDB]=3[AEB]=\frac{3}{8}[ABC]$, because of base-ratios. Using Heron's Formula on $\triangle EDB$, as it is simplest, we see that $[ABC]=3\sqrt{7}$, so your answer is $10$.