Triangle Triangle Triangle

Let $JK=x$ . Triangles $\Delta AHB$ and $\Delta JHK$ are similar, and their heights add up to $8$ . Say the perpendicular distance from $H$ to $AB$ is $h$ (the height of $\Delta AHB$ ). Since they're similar, the ratio of their heights is the same as the ratio of their bases: $\frac{8-h}{h}=\frac{x}{18}$

Rearranging, we find $h=\frac{8\times 18}{x+18}$

so the area of $\Delta AHB$ is $\Delta AHB=\frac12 \times 18 \times h = \frac{4\times 18^2}{x+18}$

In the say way, we can find $\Delta AFE=\frac{4\times 12^2}{x+12}$ and $\Delta BGE=\frac{4\times 6^2}{x+6}$

We can write the area of $EFHG$ in terms of these triangles: $[EFHG]=\Delta AHB - \Delta AFE - \Delta BGE$

that is, $10=\frac{4\times 18^2}{x+18}-\frac{4\times 12^2}{x+12}-\frac{4\times 6^2}{x+6}$

Cross-multiplying and tidying leads to a cubic in $x$ ; the only root in the interval $0<x<18$ is $x=\boxed6$ .

We note that the area of quadrilateral $EFGH$ is given by:

$[EFGH] = [JEK] - [JFK] - [JGK] + [JHK]$

Let $|JK|=x$ . Then $[JEK] = \dfrac 12 \cdot x \cdot 8 = 4x$ . For $[JFK]$ , let the altitude of $\triangle JEK$ be $h$ . Since $\triangle JFK$ and $\triangle AFE$ are similar, then $\dfrac h{8-x} = \dfrac x{12} \implies h = \dfrac {8x}{x+12}$ and $[JFK]=\dfrac {xh}2 = \dfrac {4x^2}{x+12}$ . Similarly, $[JGK] = \dfrac {4x^2}{x+6}$ and $[JHK] = \dfrac {4x^2}{x+18}$ . Then we have:

$4x - \frac {4x^2}{x+12} - \frac {4x^2}{x+6} + \frac {4x^2}{x+18} = 10$

Solving for $x$ , we get $|JK|=x=\boxed 6$ .