Notes
two squares overlapping a square solution

Solution to the Two Squares Overlapping a Square Puzzle

Solution by Congruent Triangles

In the above diagram, the lines $I K$ and $J H$ are the vertical and horizontal lines through $G$, respectively.

The angles $I \hat{G} F$ and $H \hat{G} E$ are the same, since both are equal to $90^\circ - F \hat{G} H$, and the lengths of $G E$ and $G F$ are the same, so triangles $G I F$ and $G H E$ are congruent. This means that $G H$ and $G I$ have the same length as each other, so also $G J$ and $G K$ have the same length as each other.

Also from the congruence, $I F$ and $H E$ have the same length, so the sum of the lengths of $I F$ and $E B$ is the length of $H B$ and so also of $A J$. The remainder of $D F$ is $D E$ which has the same length as $J G$. So the lengths of the red lines have the same sum as the lengths of $A J$ and $J G$.

Now $A J$ and $J G$ are equal and perpendicular meaning that $J$ is actually at the centre of the left-hand square. The quadrilateral $A K G J$ is therefore a square of area half that of the left-hand square, and so of area $9$. It therefore has side length $3$ so the sum of the red lines is $6$.

Solution by Invariance principle

The property that can vary in this diagram is the angle of the right-hand square. Drawing it so that it is oriented the same as the outer square puts it in the right-hand top corner of that square. That $D F$ and $E B$ are the sides of a square with diagonal $A G$ is then obvious.