Notes
a rectangle in a square solution

Solution to the Rectangle in a Square Puzzle

Solution by Angle at the Centre is Twice the Angle at the Circumference and Pythagoras' Theorem

In the above diagram, the line segment $O K$ is vertical and passes through $J$, while $O I$ is horizontal and passes through $H$. The line segment $D B$ is at $45^\circ$ to the horizontal and vertical, so the line segments $J F$ and $H F$ have the same length and this establishes $O H F J$ as a square. The circle has centre $O$ and passes through $H$, then since $O H$ and $O J$ have the same length, it also passes through $J$.

To see that it also passes through $A$, consider the line segment $O A$. Using Pythagoras' theorem, the square of the length of $O A$ is the same as the sum of the squares of the lengths of $E H$ and $G J$.

The blue rectangle occupies half of the area of the square. By reflecting in the line $D B$, it can be seen that the region $A E H O J G$ matches $C I H F J K$. Therefore, the central square $O H F J$ must have the same area as the two squares $E B I H$ and $G J K D$. This means that the squares of the lengths of $E H$ and $G J$ is equal to the square of the length of $O H$, and hence $O A$ has the same length as $O H$. Therefore $A$ lies on the circle.

The angle $H \hat{O} J$ is a right-angle, and $O$ is the centre of the circle, so by the result that the angle at the centre is twice the angle at the circumference, angle $H \hat{A} J$ is $45^\circ$.

Solution by Invariance Principle

The point $G$ can move providing the area of the rectangle remains half of the area of the square. It can therefore move so that it lies on the edge $C B$, at which point it must be the midpoint of that edge. The point $F$ then coincides with $B$ (and $E$) and the intersection of $D B$ with the edge $I G$ is its midpoint, which is the centre of the square. The line $A H$ is therefore the other diagonal and so angle $H \hat{A} F$ is $45^\circ$.