Notes
four squares inside a rectangle solution

Solution to the Four Squares Inside a Rectangle Puzzle

Solution by Angles in a triangle, Angles at a point on a straight line, and Angles in parallel lines

In the above diagram, $A$ is horizontally across from $F$ and $D$ is vertically below. So the quadrilateral $A C D F$ has all four interior angles $90^\circ$ and so is a rectangle.

Considering angles at $B$, and using that $F B$ is a diagonal of the left-hand square, we have:

Over at $E$, we have:

Then as triangle $B C E$ is a right-angled triangle:

Putting those together,

So triangles $F A B$ and $F D E$ are similar, and the lengths of $F B$ and $F E$ are the same, so they are congruent. This means that the lengths $F A$ and $F D$ are also the same and so the quadrilateral $A C D F$ is actually a square.

The line $C F$ is therefore at $45^\circ$ to $C D$ and so the requested angle is also $45^\circ$ by angles in parallel lines.

Solution by Invariance Principle

The squares can be drawn at any angle inside the rectangle, so they can be drawn to align with the sides of the outer rectangle. This brings $B$ to $C$ and the left-hand square sits in the lower left corner of the rectangle. The line $C F$ is then evidently at $45^\circ$ to the base of the rectangle.