Notes
four squares ix solution

Solution to the Four Squares IX Puzzle

Solution by Similar Triangles and Properties of Isosceles Right-angled Triangles

In the above diagram, line segment $D H$ is the continuation of $E D$, and $G$ is such that angle $F \hat{G} D$ is $90^\circ$.

Angles $A \hat{C} B$, $E \hat{C} A$, and $D \hat{C} E$ add up to $180^\circ$ since they are angles at a point on a straight line, therefore angles $D \hat{C} E$ and $A \hat{C} B$ add up to $90^\circ$, since angle $E \hat{C} A$ is $90^\circ$. Then also angles $A \hat{C} B$ and $B \hat{A} C$ add up to $90^\circ$ since $A B C$ is a right-angled triangle. Hence angles $D \hat{C} E$ and $B \hat{A} C$ are the same, so triangles $A B C$ and $C D E$ are similar.

Since line segment $A C$ has twice the length as $C E$, so also $A B$ has twice the length of $C D$. But then since $B D$ has the same length as $A B$, this means that $A B$ also has twice the length of $B C$.

Angles $G \hat{E} F$ and $C \hat{E} D$ add up to $90^\circ$, so triangles $C D E$ and $E G F$ are likewise similar, but then as line segments $C E$ and $E F$ are sides of the same square, they have the same length. So in fact, triangles $C D E$ and $E G F$ are congruent.

This means that line segment $E G$ has twice the length of both line segments $F G$ and $E D$. Therefore, line segments $D G$ and $F G$ have the same length, so triangle $F G D$ is an isosceles right-angled triangle.

Therefore, angle $G \hat{D} F$ is $45^\circ$, so angle $C \hat{D} F$ is $90^\circ + 45^\circ = 135^\circ$.