Notes
three congruent squares overlapping a square solution

Solution to the Three Congruent Squares Overlapping a Square Puzzle

Solution by Isosceles Triangles, Angles at a Point, Angles at a Point on a Straight Line, and Area Scale Factor

The first step is to establish that point $B$ lies on the diagonal $A C$.

Consider triangles $A F E$ and $E D C$. These are congruent as the sides match in length. Therefore angles $E \hat{F} G$ and $B \hat{D} C$ are equal, so triangles $G F E$ and $B D C$ are also congruent to each other. By considering the angles at $F$, since angles at a point add up to $360^\circ$, angles $B \hat{D} C$ and $A \hat{F} B$ add up to $180^\circ$. So using angles in isosceles triangles, angles $A \hat{B} F$ and $C \hat{B} D$ add up to $90^\circ$. Therefore angle $A \hat{B} C$ is $180^\circ$ so $A B C$ is a straight line.

As the total area of the three squares is equal to the area of the large square, the length scale factor from the small to the large is $\sqrt{3}$. This means that $E H$ has length $\frac{3}{2}$ times that of $E B$. As it is a right-angled triangle, this establishes that $E H B$ is half an equilateral triangle. By symmetry, so is triangle $H I B$ and so angle $B \hat{I} H$ is $30^\circ$. Angle $B \hat{I} A$ is therefore $75^\circ$.