Notes
three squares vi solution

Solution to the Three Squares VI Puzzle

Solution by Pythagoras' Theorem and Angles in a Quadrilateral

Let the side lengths of the squares be $a$, $b$, and $c$ in increasing order.

Then $E B$ has length $a + b - c$ so the green region has area $a b + b^2 - c b$.

The line segment $F C$ has length $c - a$, so the blue region has area:

These areas are the same, so:

This means that the lengths $a$, $b$, and $c$ form the sides of a right-angled triangle, by the converse to Pythagoras' theorem. Triangle $A B C$ is a right-angled triangle with sides $a$ and $b$, so its hypotenuse, $A B$, has length $c$.

This means that triangles $A D B$ and $F A B$ are both isosceles with angles $A \hat{D} B$ and $B \hat{A} D$ equal to each other, and $B \hat{F} A$ and $F \hat{A} B$ also equal to each other.

Then as the angles in a quadrilateral add up to $360^\circ$:

Hence the angle is $135^\circ$.