<s>[INST] In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$. 
In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$. If $CD = 7$ and $[\Delta ABD] = a\sqrt{5}$ , find $a$ .

What I Tried: Here is a picture:-

Let the perpendicular bisector of $BC$ pass through $BC$ at $E$ . 
Then I first noticed that $\Delta BDE \cong \Delta CDE$ from $(SAS)$ congruency. 
This gives the required information in the diagram, as well as we have $BD = 7$ . 
Now :- $$\Delta ABD \sim \Delta ACB$$
$$\rightarrow \frac{AD}{AB} = \frac{7}{BC} = \frac{AB}{AC}$$
So let $AD = k$ , $AB = m$ , $BE = EC = n$ . We have :-
$$\frac{k}{m} = \frac{7}{2n} = \frac{m}{(7+k)}$$
 [/INST] Let $E=(0,0)$, $B=(-\alpha,0)$, $C=(\alpha,0)$, $D=(0,\beta)$. Then we have
$$\alpha^2+\beta^2=7^2=49,\quad\tan \angle ACB=\frac{\beta}{\alpha}\in(0,\sqrt3).$$
Let $t=\tan \angle ACB$, then
$$y_{AB}=\frac{2t}{1-t^2}(x+\alpha),\quad y_{AC}=-tx+\beta.$$
And we have
$$A=\left(-\alpha\frac{1+t^2}{3-t^2},\alpha\frac{4t}{3-t^2}\right),\quad |AB|=2\alpha\frac{1+t^2}{3-t^2}.$$
Thus, the area
\begin{align}
S_{\triangle ABD} &= \frac12|AB|\cdot|DE|\\
&=\alpha\beta\frac{1+t^2}{3-t^2}\\
&=\frac{49t}{3-t^2}\in(0,\infty).
\end{align}
If you have some other conditions such that $\frac{t}{3-t^2}=\sqrt5$, then $a=49$. If not, $a$ can be any positive real number.
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