The figure shows a square ABCD. EH = EF = EG. Find $\angle\text{FGH}$. Answer : ___

1. Identify the lengths of $EG$ and $HF$ in terms of the square's side length.

Let the side length of the square $ABCD$ be $s$.
Since $EG$ is a vertical line segment from $AB$ to $DC$, $EG = s$.
Since $HF$ is a horizontal line segment from $AD$ to $BC$, $HF = s$.
We are given $EH = EF = EG$, so $EH = EF = s$.

2. Find the lengths of $HI$ and $EI$.

The vertical line $EG$ and horizontal line $HF$ intersect at right angles at $I$. By symmetry (since $EH = EF$), $I$ is the midpoint of $HF$.
Thus, $HI = \frac{HF}{2} = \frac{s}{2}$.
In the right-angled $\triangle EIH$, using Pythagoras' theorem:
$EI = \sqrt{EH^2 - HI^2} = \sqrt{s^2 - (\frac{s}{2})^2} = \sqrt{\frac{3s^2}{4}} = \frac{s\sqrt{3}}{2}$.

3. Calculate $\angle HGI$ and hence $\angle FGH$.

The length of $IG$ is $EG - EI = s - \frac{s\sqrt{3}}{2} = s(1 - \frac{\sqrt{3}}{2})$.
In the right-angled $\triangle HIG$:
$\tan(\angle HGI) = \frac{HI}{IG} = \frac{\frac{s}{2}}{s(1 - \frac{\sqrt{3}}{2})} = \frac{\frac{1}{2}}{1 - \frac{\sqrt{3}}{2}} = \frac{1}{2 - \sqrt{3}}$.
Rationalising the denominator:
$\frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = 2 + \sqrt{3}$.
$\angle HGI = \tan^{-1}(2 + \sqrt{3}) = 75^\circ$.
By symmetry, $\angle FGI = \angle HGI = 75^\circ$.
Therefore, $\angle FGH = 75^\circ + 75^\circ = 150^\circ$.