Notes
two overlapping squares ii solution

Solution to the Two Overlapping Squares II Puzzle

Solution by Similar Triangles, Congruency, Angles in a Square, and Vertically Opposite Angles

In the extended diagram above, the line $E H$ is the continuation of $F E$ until it meets the extension of $C D$. The point $I$ makes $A D H I$ into a rectangle, and $J$ is so that $I J$ is perpendicular to $K H$.

Since $F E H$ is a straight line, angle $D \hat{E} H$ is $90^\circ$. Angles $H \hat{D} E$ and $A \hat{D} G$ are equal as both combine with angle $E \hat{D} A$ to form a right-angle, and line segments $G D$ and $E D$ have the same length. These establish triangles $A G D$ and $E D H$ as congruent so $D H$ has the same length as $A D$ which means that $A D H I$ is actually a square and is the congruent to $A B C D$.

Each side of triangles $A G D$ is parallel to one of the sides of triangle $I J H$ and sides $A D$ and $I H$ are the same length so triangles $A G D$ and $I J H$ are congruent. This means that the region $I A D H J$ is congruent to the yellow region.

The yellow region comprises half the area of the original diagram, meaning that the orange and white make up the other half. Since triangles $A G D$ and $E H D$ are congruent, the quadrilateral $A D H F$ therefore has the same area as the yellow region, which in turn has the same area as $I A D H J$. Therefore, triangles $A F K$ and $I K J$ have the same area. These triangles are similar, since both are right-angled and angles $F \hat{K} A$ and $J \hat{K} I$ are equal as they are vertically opposite angles. Therefore they are congruent and in particular, $A F$ and $J I$ have the same length. This establishes $A$ as the midpoint of $F G$.

Triangle $A G D$ therefore is one quarter of the area of the square $F G D E$ and so the orange region is $\frac{3}{4}$ of that square, which is itself half the total area. So the area of the orange region is $\frac{3}{8}$ths of the total area.