# Solution to the Two Overlapping Squares II Puzzle +-- {.image} [[TwoOverlappingSquaresII.png:pic]] > Half the total area of these overlapping squares is yellow. What fraction is orange? =-- ## Solution by [[Similar Triangles]], [[Congruency]], [[Angles in a Square]], and [[Vertically Opposite Angles]] +-- {.image} [[TwoOverlappingSquaresIILabelled.png:pic]] =-- 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 triangle|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.