Proportional square

Solution:

If $\square ABCD$ 's side's length is $a$ , then points $A$ , $B$ , $C$ and $D$ are $(0, 0)$ , $(a, 0)$ , $(a, a)$ and $(0, a)$ , respectively. We will draw a segment $FB$ , that is parallel to the $AG$ and that the point $F$ lies on side $AD$ (on extension). A point $H$ shall be the midpoint of the segment $FB$ . Segments $EH$ and $AG$ have a common point $T$ . According to Thales's theorem $\frac{HT}{HE} = \frac{BX}{BE}$ . Point $F(0, -\frac{a}{4})$ , therefore point $H(\frac{a}{2}, -\frac{a}{8})$ . Similarly, $T(\frac{a}{2}, \frac{a}{8})$ . Therefore, we will get that $HT$ is $\frac{2a}{8}$ and $HE$ is $\frac{9a}{8}$ . Then $\frac{HT}{HE} = \frac{BX}{BE} = \frac{2}{9}$ .

The answer is $11$ .