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.
