How to find $AL$ in a triangle with one side trisected? Here is the original diagram of the question:

Let $D$ and $E$ be the trisection points of $BC$. And let $K$ and $L$ be points on $AB$ and $AC$ such that $\angle BKE= \angle DLC = \alpha$. If $KB=16$, $AK=12$ and $LC=6$. Then find the length of $AL$.
The only thing I could think of that could use the fact that $\angle BKE$ and $\angle DLC$ are equal is the sine law. Also for convenience let $BD=DE=EC=y$.
$$\dfrac{2y}{\sin \alpha}=\dfrac{16}{\sin (\alpha + \angle B)}=\dfrac{KE}{\sin\angle B}$$
$$\dfrac{2y}{\sin \alpha}=\dfrac{6}{\sin{(\alpha + \angle C)}}=\dfrac{LD}{\sin \angle C}$$
Since I got $\sin \angle C$ and $\sin \angle B$ I thought of using sine rule again in the original triangle.
$$\dfrac{x+6}{\sin \angle B}=\dfrac{28}{\sin \angle C}$$
$$\begin{align*}
\dfrac{\sin (\alpha + \angle B)}{\sin ( \alpha + \angle C)} & =\dfrac{16}{6} \\
\implies \dfrac{\sin (\alpha + \angle B)+\sin ( \alpha + \angle C)}{\sin (\alpha + \angle B)-\sin ( \alpha + \angle C)} & = \dfrac{11}{5} \\
\implies 2 \cdot \tan \biggl(\alpha + \dfrac{B+C}{2} \biggr) \cot \biggl( \dfrac{B-C}{2} \biggr) & = \dfrac{11}{5}
\end{align*}$$
But I could not really get anywhere with these equations. Since the terms got very messy and I really could not simplify anything.
Would anyone please provide an elementary solution? Or perhaps give a hint on how to complete my approach?

Extends $LD$ to meet $AB$ at $F$, extends $KE$ meets $AC$ at $G$.
Let $L'$ be on $FL$ such that $BL'\mathbin{\!/\mkern-5mu/\!} AL$. Similarly $CK'\mathbin{\!/\mkern-5mu/\!}AK$.
Then $BL'=\frac 12CL=3$, $CK'=\frac 12BK=8$.
Now since triangles $CK'G$ and $AKG$ are similar, we have $AG=AC\cdot \frac{AK}{AK-CK'}=3(x+6)$. Similarly $AF=28\frac x{x-3}$.
Now since triangles $AFL$ and $AGK$ are similar, we have $\frac {AF}{AL}=\frac {AG}{AK}$, then:
\begin{align}
\frac {28x}{x(x-3)}&=\frac {3(x+6)}{12}\\\\
4\cdot 28&=(x-3)(x+6)\\\\
112&=x^2+3x-18\\\\
0&=(x-10)(x+13)\\
\end{align}
Giving us $x=10$.
