Notes
three squares and two triangles solution

Solution to the Three Squares and Two Triangles Puzzle

Solution by Similar Triangles, Area of a Triangle, and Parallel Lines

With the points labelled as above, consider triangles $A F C$ and $A E C$. Triangle $A F C$ consists of triangles $A G C$ and $A F G$, while triangle $A E C$ consists of triangles $A G C$ and $G E C$. Since triangles $A F G$ and $G E C$ have the same area, so also do triangles $A F C$ and $A E C$. Taking $A C$ as the base of each, their heights above $A C$ must therefore be the same. This means that the perpendicular distance from $F$ to $A C$ is the same as that from $E$, so $F E$ is parallel to $A C$.

Since $F C$ is also parallel to $A B$, this means that triangles $F E H$ and $A C B$ are similar. Therefore, the ratio of the length of $E H$ to that of $F H$ is $1 : 2$. Since $E H$ has the same length as $H C$, this means that $H$ divides $F C$ in the ratio $2 : 1$ and so the side length of the smaller square is two thirds that of the larger.

Therefore, the length of $E H$ is two thirds that of $A F$, and so since triangles $A F G$ and $G E C$ have the same area, the length of $F G$ must be two thirds that of $G C$. The ratio of $F G$ to $G C$ is therefore $\frac{2}{3} : 1 = 2 : 3$. The area of triangle $G C A$ is therefore $\frac{3}{2}$ times that of $F G A$, hence is $\frac{27}{2}$.

The area of triangle $F C A$ is then $9 + \frac{27}{2} = \frac{45}{2}$, so the area of the two larger squares is $45$.

The smaller square has side length $\frac{2}{3}$ that of the larger, so its area is $\frac{4}{9}$ of the larger, which is $\frac{4}{9} \times \frac{45}{2} = 10$.

The area of the three squares is therefore $45 + 10 = 55$.