Lots of ratio

Let $ST : TR = 1 : n$

Then $\Large\frac{PU}{UT}$ $= \Large\frac{ar(\triangle PUQ)}{ar(\triangle QUT)}$

And $\Large\frac{PU}{UT}$ $= \Large\frac{ar(\triangle PUS)}{ar(\triangle SUT)}$

Using the above two equation we get

$\Large\frac{PU}{UT}$ $= \Large\frac{ar(\triangle PQS)}{ar(\triangle SQT)}$ $= \Large\frac{\frac{1}{2}ar(PQRS)}{\frac{1}{n+1}ar(\triangle SQR)}$ $= \Large\frac{\frac{1}{2}ar(PQRS)}{\frac{1}{n+1}\frac{1}{2}ar(PQRS)}$ $= n+1$

With reference to the diagram of Chew-Seong Cheong's solution

Let $AB = a$ and it is given $AD = 12$ . Then clearly $EE' = \Large\frac{a}{2}$ and $AE = 6$

Now applying the concept shown above we will get

$\Large\frac{AF'}{F'E'}$ $= 1 + 1 = 2$

So, $\Large\frac{AF}{FE}$ $= \Large\frac{AF'}{F'E'}$ $= 2$

$\Rightarrow AF = 4$

Applying that concept again

$\Large\frac{AG'}{G'F'}$ $= 2+ 1 = 3$

So, $\Large\frac{AG}{GF}$ $= \Large\frac{AG'}{G'F'}$ $= 3$

$\Rightarrow AG = 3$

Let $D$ be the origin $(0,0)$ of an $xy$ -plane. The length of $AB=CD=a$ . Let the other ends of the three short parallel lines be $E'(x_1a,y_1)$ , $F'(x_2a,y_2)$ , and $G'(x_3a,y_3)$ . Let us find the recursive relation of $\{x_n, y_n\}$ . First consider a general case of $(x_2,y_2)$ .

Let $E'E''$ and $F'F''$ be perpendicular to $CD$ and $EE'$ respectively. Due to similar triangles we have:

$\begin{cases} \dfrac {EF''}{F'F''} = \dfrac {EB'}{BB'} & \implies \dfrac {x_2}{y_2-y_1} = \dfrac 1{12-y_1} & ...(1) \\ \dfrac {F''E'}{F'F''} = \dfrac {CD}{AB} & \implies \dfrac {x_1-x_2}{y_2-y_1} = \dfrac 1{12} & ...(2) \end{cases}$

$\begin{aligned} \dfrac {(1)}{(2)}: \quad \frac {x_2}{x_1-x_2} & = \frac {12}{12-y_1} \implies x_2 & = \frac {12x_1}{24-y_1} & \implies x_n = \frac {12x_{n-1}}{24-y_{n-1}} \end{aligned}$

$\begin{aligned} (2): \implies y_2 & = 12(x_1 - x_2) + y_1 & \implies y_n = 12(x_{n-1}-x_n) + y_{n-1} \end{aligned}$

We can easily find that $x_1 = \dfrac 12$ and $y_1 = 6$ , then

$\begin{array} {ll} x_2 = \dfrac {12 \times \frac 12}{24 - 6} = \dfrac 13 & \implies y_2 = 12\left(\dfrac 12 - \dfrac 13\right) + 6 = 8 \\ x_3 = \dfrac {12 \times \frac 13}{24 - 8} = \dfrac 14 & \implies y_3 = 12\left(\dfrac 13 - \dfrac 14\right) + 8 = 9 \end{array}$

We note that $AG = AD-GD = 12 - y_3 = \boxed 3$ .

Let the position coordinates of $A, B, C, D$ be $(0,12),(a, 12),(a, 0),(0,0)$ respectively. Then those of $E$ are $(0,6)$ and the equation of $\overline {AC}$ is

$y=-\dfrac{12}{a}x+12$ .

Equation of $\overline {BE}$ is

$y=\dfrac{6}{a}x+6$ .

Solving, the $y$ -coordinate of $F$ is $8$ , and the equation of $\overline {BF}$ is

$y=\dfrac{4}{a}x+8$ .

So, the $y$ -coordinate of $G$ is $9$ and $|\overline {AG}|=12-9=\boxed 3$ .