Notes
four squares xii solution

Solution to the Four Squares XII Puzzle

Solution by Properties of Isosceles Triangles, Angles in a Triangle, and Angles at a Point on a Straight Line

With the points labelled as above, consider triangles $B C I$ and $I H J$. Both are right-angled. The sides $B C$ and $I H$ are sides of the purple squares and so have the same length. The sides $C I$ and $H J$ are the difference between the side of the outer square and one purple square, so have the same length. Therefore, triangles $B C I$ and $I H J$ are congruent so triangle $B I J$ is isosceles.

Angles $B \hat{I} C$ and $H \hat{I} J$ add up to $90^\circ$ since the angles in a triangle add up to $180^\circ$, so since angles at a point on a straight line add up to $90^\circ$, angle $J \hat{I} C$ is $90^\circ$.

Therefore, angle $B \hat{J} I$ is half of $180^\circ - 90^\circ = 90^\circ$, so is $45^\circ$.

Solution by Invariance Principle

The size of the outer square can vary, so drawing it as small as possible produces the configuration as above, from which the angle is clearly $45^\circ$.