Let there be a $\triangle ABC$, such that $AB = 5$, $BC = 6$, and $AC = 4$. The bisectors of angles $B$ and $C$ intersect at point $I$. Draw a parallel to $BC$ through $I$, intersecting $AC$ and $AB$ at $E$ and $D$, respectively. What is the value of $ED$?
(Ans:$\frac{18}{5}$)
I tried as follows.
$$\frac{5}{AK} = \frac{6}{4-AK} \implies 20-5AK = 6AK \therefore AK = \frac{20}{11} \implies CK = \frac{24}{11}$$
$$\frac{4}{AL} = \frac{6}{5-AL} \implies 20-4AL = 6AL \therefore AL = AL = 2 \implies BL = 3$$
$$\triangle ADE \sim \triangle ACB: \frac{DE}{6}=\frac{AE}{5}=\frac{AD}{4}\tag I$$
$$\triangle AED: \frac{AE}{AD} = \frac{EI}{DI}$$
from $(I)$, $$\frac{EI}{DI} = \frac{5}{4}$$
Unfortunately, I don't offhand know how to finish using what you've done. Instead, here's a different method. As indicated in the diagram below, add the line $IF$, and the perpendicular line from $BC$ that goes through $A$, with it crossing $ED$ at $M$ and $BC$ at $N$. Let $\lvert IF\rvert = r$ (so $\lvert MN\rvert = r$ also) and $\lvert AM\rvert = h$.
The incircle radius relation to the triangle area, using that $a$, $b$ and $c$ are the side lengths of $\triangle ABC$, is
$$\Delta = sr, \;\;\; s = \frac{a+b+c}{2} = \frac{15}{2} \tag{1}\label{eq1A}$$
Using $BC$ as the base, with the one-half base length times the height (i.e., $\lvert AN\rvert$) triangle area formula, we get
$$sr = \left(\frac{1}{2}\right)6(r + h) \;\;\to\;\; r + h = \frac{sr}{3} \;\;\to\;\; h = \frac{(s - 3)r}{3} \tag{2}\label{eq2A}$$
Since $DE \parallel BC$, then $\triangle ADE \sim \triangle ACB$ so, using \eqref{eq1A} and \eqref{eq2A}, gives that
$$\begin{equation}\begin{aligned} \frac{\lvert ED\rvert}{\lvert BC\rvert} & = \frac{\lvert AM\rvert}{\lvert AN\rvert} \\ \frac{\lvert ED\rvert}{6} & = \frac{\frac{(s-3)r}{3}}{\frac{sr}{3}} \\ \lvert ED\rvert & = \frac{6(s-3)}{s} \\ \lvert ED\rvert & = \frac{18}{5} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$