no trig

The first step in order to tackle this problem is to draw another square enclosing square DEFG.

First, we prove that $WXYZ$ is actually a square: 

$\because\overline{ED}=\overline{DG}=\overline{GF}=\overline{FE}\because\angle{Z}=\angle{Y}=\angle{YXW}=\angle{ZWX}\because\angle{DEZ}+\angle{WEF}=90º,\angle{DEZ}+\angle{ZDE}=90º\Rightarrow\angle{ZDE}=\angle{WEF}.$

$\text{Using the same reasoning, we get:} \angle{ZDE}=\angle{WEF}=\angle{DGY}=\angle{GFX}.$
$\therefore\text{By AAS congruency:} \triangle{ZDE}\cong\triangle{YGD}\cong\triangle{XFG}\cong\triangle{WEF}.$

From this, we get

$\overline{ZE}+\overline{EW}=\overline{ZD}+\overline{DY}=\overline{YG}+\overline{GX}=\overline{FX}+\overline{FW},$which simplifies to $\overline{ZW}=\overline{ZY}=\overline{YX}=\overline{XW}.$

Therefore $WXYZ$ is a square.

Since $\triangle ABC$ is equilateral,$\angle B=60º. \because \triangle BEW$ is a 30-60-90 triangle, $\frac{\overline{EW}}{\overline{BW}}=\sqrt3.  \text{Same goes with } \triangle GXC, \frac{\overline{GX}}{\overline{XC}}=\sqrt3.$
$\text{If }\overline{EW}=x \text{ and } \overline{GX}=y,\text{ we get }\overline{BW}=\frac{x}{\sqrt3} \text{ and } \overline{XC}=\frac{y}{\sqrt3}.$

If the equilateral triangle's side length is $a, a=\overline{BW}+\overline{WF}+\overline{FX}+\overline{XC}=\frac{x}{\sqrt3}+y+x+\frac{y}{\sqrt3}.$

After simplifying, we get $x+y=\frac{3-\sqrt3}{2}a.$

$\because \overline{WX}=x+y \therefore x+y=\overline{DH}.$

Since in any case, $\overline{DH}=\frac{3-\sqrt3}{2}a,$ the length remains consistent.