Notes
seven squares solution

Solution to the Seven Squares Puzzle

Solution by Angle in a Semi-Circle and Angle at the Circumference is Half the Angle at the Centre

With the points labelled as above, consider a circle centred on $O$ which passes through $A$. Since the pink squares are all the same size, $O A$, $O B$, and $O C$ all have the same length and therefore the circle also passes through $B$ and $C$.

Triangle $A D C$ is a right-angled, since $D$ is the corner of the purple square. Therefore, as the angle in a semi-circle is a right-angle, $D$ also lies on the same circle.

Then as the angle at the circumference is half the angle at the centre, angle $A \hat{D} B$ is half of angle $A \hat{O} B$, but this is the corner of a square and so is $90^\circ$.

Therefore, angle $A \hat{D} B$ is $45^\circ$.

Solution by Isosceles Triangles and Angles in a Quadrilateral

In the above diagram, $E$ and $F$ are so that $O E$ and $O F$ are perpendicular to $A D$ and $D C$ respectively. Triangles $O E A$ and $C F O$ are congruent because $C O$ and $O A$ have the same length, so $E$ is at the midpoint of $A D$. This means that triangle $O E D$ is also congruent to $O E A$, so $O D$ also has the same length as $O A$.

Hence triangles $O D A$ and $O D B$ are each isosceles. So the sum of angles $D \hat{B} O$ and $O \hat{A} D$ is equal to angle $A \hat{D} B$, meaning that the sum of the angles inside quadrilateral $A D B O$ is $2 A \hat{D} B + B \hat{O} A$, where the latter is taken inside the quadrilateral. That angle is $360^\circ - 90^\circ = 270^\circ$ and so, since the angles in a quadrilateral add up to $360^\circ$, angle $A \hat{D} B$ is $\frac{360^\circ - 270^\circ}{2} = 45^\circ$.

Solution by Invariance Principle

The larger square can tilt in relation to the arrangement of smaller squares, giving configurations where the answer can be more simply deduced.

In this configuration, point $D$ aligns with $C$, meaning that angle $A \hat{D} B$ coincides with the diagonal in the upper left square, whence is $45^\circ$.

In this configuration, $D$ is such that $B O D$ is a straight line and the corners of the large square are at $A$, $B$, $C$, and $D$, whence angle $A \hat{D} O = 45^\circ$.