Inequality in a triangle: A line segment dividing opposite side with ratio 3:1 
In triangle $\triangle ABC$, $D$ is a point on side $BC$ such that $|DC|=3 |BD|$. Also $|AB|=24$ and $|AC|=18$. Find the interval of all possible values of length of line segment $AD$. 


I wrote $|DC|=3k$ and $|BD|=k$. Triangle inequality gives 
$$ 24-18<4k<24+18\ \ \ \Longrightarrow 1.5<k<10.5$$
Then I used Stewart theorem to express $x$ in terms of $k$. The relation was $$x^2=513-3k^2$$Which gave me $13.5<x<22.5$.
Can we find a solution without Stewart theorem (and without trigonometry). I feel that there must be an easier (or more basic) solution around.

We can use the Heron's formula:
$$S_{\Delta ABC}=\frac{1}{4}\sqrt{\sum_{cyc}(2a^2b^2-a^4)}.$$
Now, let $BD=y$, $DC=3y$ and since $S_{\Delta ADC}=3S_{\Delta ABD},$ we obtain:
$$2(18^2x^2+54^2y^2+9x^2y^2)-18^4-x^4-81y^4=9(2(24^2x^2+24^2y^2+x^2y^2)-24^4-x^4-y^4)$$ or
$$(x^2+3y^2-513)(x^2-3y^2-702)=0$$ and since $x^2=3y^2+702$ is impossible, we obtain:
$$x^2+3y^2=513.$$
Can you end it now?
