Notes
two squares and an equilateral triangle solution

Two Squares and an Equilateral Triangle

Solution by Equations of Lines

Position the diagram on a coordinate grid so that point $A$ is at the origin and point $C$ at the point $(1,0)$. From the properties of equilateral triangles, the coordinates of $G$ can be seen to be $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and so the equation of the line extending $C G$ is:

Point $F$ lies on this line and also on the line $y = x$. Its coordinates can therefore be calculated to be $\left( \frac{\sqrt{3}}{1 + \sqrt{3}}, \frac{\sqrt{3}}{1 + \sqrt{3}}\right)$.

The length of $B C$ is then $1 - \frac{\sqrt{3}}{1 + \sqrt{3}} = \frac{1}{1 + \sqrt{3}}$ and so the coordinates of $D$ are $\left(1, \frac{1}{1 + \sqrt{3}}\right)$. The length of $G H$ is therefore:

The triangle $G H D$ is therefore an isosceles right-angled triangle and so angle $G \hat{D} H$ is $45^\circ$.