Dudeney’s solutions to haberdasher's problem exact measures of sections What is the IG length if the side of the square is 1? I wonder if it is half of the square side. The triangle below represents the haberdasher's problem. 
version 2

version 1 (added after edit, here the question is about JL)

The images are taken from here.
Edit. I posted two versions of the Dudeney triangle to square dissection. Original question was intended to concern the proper Dudeney dissection. It turned out that the version 2 is a dissection of equilateral triangle to rectangle, not a square. Though the rectangle is almost a square. Special thanks to SMM for revealing the truth.
 A: The upper side of the square is formed by $IG_{red}+IG_{green}=2IG$, because (as you can see below) the square is formed from the triangle by rotating the green piece by 180° about $G$, rotating the blue piece by 180° about $F$, and finally translating the yellow piece in place.
Hence you are right: IG is a half of the square side.

EDIT.
I copied below Dudeney's original construction (the second construction on that site is a fake). Suppose the side of the equilateral triangle is $2$, so that its altitude and its area are both $\sqrt3$. The side of the square is then $\root{4}\of3$.
In the construction (warning: the names of some points are different from those in the above diagram), $D$ and $E$ are midpoints of $AB$ and $BC$, while altitude $AE$ is produced to $F$ with $EF=EB=1$. It follows that the radius $AG$ of circle $AHF$ is $(\sqrt3+1)/2$. $BE$ is then produced to meet the circle at $H$. In right triangle $HEG$ we know that $GH=(\sqrt3+1)/2$ and $EG=AE-AG=(\sqrt3-1)/2$. From Pythagoras' theorem it follows then $EH=\root{4}\of3$, hence $EH$ is the same as the side of the square we want to construct.
Circle $HI$ is centered at $E$, hence $EI=EH=\root{4}\of3$. Let now $\alpha=\angle EIC$: by sine rule applied to triangle $EIC$ we have 
$$
\sin\alpha={\sin60°\cdot EC\over EI}={\root{4}\of3\over2}.
$$
Point $J$ is then chosen such that $IJ=EC=1$ and $JL$ is perpendicular to $EI$. Hence 
$$JL=IJ\sin\alpha={\root{4}\of3\over2}$$ 
is exactly half side of the square.

You can read below a part of the article where Martin Gardner presented this problem: the construction given there is the same shown above. Gardner, in an addendum, also warns the reader that points $I$ and $J$ (figure above) are not exactly below $D$ and $E$.

A: Denote by $a$ the side of the triangle. Since the areas of the triangle and the square are the same, we have $\frac{a^2\sqrt 3}{4}=1$, so $a=\frac{2}{\sqrt[4]3}$.
Set the coordinate system as follows: $A=(0,0)$, $C=(a,0)$ and $B=(\frac{a}{2},\frac{a\sqrt 3}{2})$. Then $D=\frac{A+B}{2}=(\frac{a}{4},\frac{a\sqrt 3}{4})$ and $E=\frac{C+B}{2}=(\frac{3a}{4},\frac{a\sqrt 3}{4})$. Further, by projecting $D$ and $E$, $F=(\frac{a}{4},0)$ and $G=(\frac{3a}{4},0)$.
We have $\vec{FE}=(\frac{a}{2},\frac{a\sqrt 3}{4})$, and $I=F+\alpha\vec{FE}=(\frac{a}{4}+\alpha\frac{a}{2},\alpha\frac{a\sqrt 3}{4})$ for some $\alpha$. Then $\vec{GI}= (\alpha\frac{a}{2}-\frac{a}{2},\alpha\frac{a\sqrt 3}{4})$. Since $FE\perp GI$ we have $\vec{FE}\cdot\vec{GI}=0$, so $\frac{a}{2}(\alpha\frac{a}{2}-\frac{a}{2})+ \frac{a\sqrt 3}{4}\alpha\frac{a\sqrt 3}{4}=0$, i.e. $\alpha-1+\frac{3}{4}\alpha=0$, where from $\alpha=\frac{4}{7}$.
So $\vec{GI}=(\frac{2a}{7}-\frac{a}{2},\frac{a\sqrt 3}{7})$. Therefore $GI^2=(\frac{2a}{7}-\frac{a}{2})^2+(\frac{a\sqrt 3}{7})^2= \frac{4a^2}{49}-\frac{2a^2}{7}+\frac{a^2}{4}+\frac{3a^2}{49}= \frac{a^2}{4}-\frac{a^2}{7}=\frac{3a^2}{28}$. Thus $GI=a\frac{\sqrt 3}{2\sqrt 7}= \frac{\sqrt[4]{3}}{\sqrt 7}$, and this is not $\frac{1}{2}$.

Edit. After Aretino posted his solution, which is correct, I started to wonder where is the mistake in mine. And there is no mistake. Mistake is in Version 2 of the construction. Namely, the rectangle obtained by Version 2 is not a square. 
