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.
 A: Idea: $\phi = \angle BDA $ and $BD = t$ then 
$$ {x^2+t^2-24^2\over 2tx}= \cos \phi  = -{x^2+9t^2-18^2\over 6tx}$$
So $$ 3x^2+3t^2-3\cdot 24^2 = -x^2-9t ^2+18^2$$
so $$ x^2+3t^2= 513$$
Also $x+t> 24$ and $ x+3t>18$
By Cauchy inequality we have $$ (1+{1\over 3})(x^2+3t^2)\geq (x+t)^2$$
so we have $x+t\leq \sqrt{684} <27$
A: Minimum value of $AD$ is when $ABC$ are aligned, with $C$ between $A$ and $B$. In that case $BC=6$ and $BD=1.5$, hence $AD=24-1.5=22.5$.
Maximum value of $AD$ is when $ABC$ are aligned, with $A$ between $B$ and $C$. In that case $BC=42$ and $BD=10.5$, hence $AD=24-10.5=13.5$.
That's because $D$ is connected to $C$ by a homothety of center $B$ and ratio $1/4$. Hence $D$ describes a circle whose center has a distance from $B$ of $24/4=6$, and radius $18/4=4.5$.

A: Using triangle inequality, $6<BC<42$. From here we can conclude that $1.5<BD<10.5$.  From $\triangle{ABD}$ we have that $13.5<x<22.5$ ($x$ is minimal when $BC$ is maximum and vice versa). 
A: 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?
