Trisected Triangle Perimeter ICTM State 2017 Division AA Precalculus Individual Question 17 reads:
$C$ and $D$ lie on $BE$ such that $AC$ and $AD$ trisect $\angle$BAE in $△ABE$. $BC = 2$, $CD = 3$, and $DE = 6$. Then the perimeter of $△ABC$ may be expressed as $P = f + k \sqrt{w} + p \sqrt{q}$ in simplified and reduced radical form. Determine the sum $(f+k+w+p+q)$. See figure below

What I tried.
I tried to create a system of equations equating the angles of the three triangles to 180°. I also equated the supplementary angles.
There were not enough equations to solve for every variable.

Let: $b=AB$, $c=AC$, $d=AD$, $e=AE$, $\alpha=\angle BAC=\angle CAD=\angle DAE$.
From the bisector theorem we get: $e=2c$ and $d={3\over2}b$. From the cosine law we then obtain:
$$
b^2+c^2-2bc\cos\alpha=4,\quad
c^2+{9\over4}b^2-3bc\cos\alpha=9,\quad
{9\over4}b^2+4c^2-6bc\cos\alpha=36.
$$
We can easily eliminate $bc\cos\alpha$ and
solve for $b^2$ and $c^2$, obtaining $b=2\sqrt{10}$, $c=3\sqrt6$. Hence $e=6\sqrt6$ and the perimeter can be computed.
