Congruency and parallelism in a triangle

Question

For the following example, one of the solutions says $\frac{6-x}{6} = \frac{x}{18}$ if one of the side of the square is called as $x$. I don't understand how the proportion and the equality are established. Could you picture?

Answer 1

Here is the solution,

$$\frac{AN}{NB}=\frac{HK}{KB}=\frac{AX}{XH}$$

Where, $X$ is the point of intersection of $MN$ and $AH$.

Therefore, with $HK = x/2$, $KB = 9 - x/2$, $AX = 6-x$, and $XH = x$

$$\frac{x/2}{9-x/2}=\frac{6-x}{x}$$

With some componendo-dividendo kind of things, you get, $$\frac{x}{18}=\frac{6-x}{6}$$

Answer 2

$\rm\color{blue} {\triangle} ANM$ and$\triangle ABC $ are similar. because $NM$ and $BC$ are parallel. so$\frac{6-x}{x}=\frac{6}{18}$. Now solve for $x$