Notes
three squares ii solution

Solution to the Three Squares (II) Puzzle

Solution by Calculating Angles

The key to this solution is to spot the right-angled triangles. Let us label the vertices as in the following diagram.

Then triangle $A D C$ is a right-angled triangle, as is triangle $A E C$. In both of these then $A C$ is the hypotenuse. Marking the midpoint of the hypotenuse and joining it to the vertex with the right-angle splits a right-angled triangle into two isosceles triangles (see right-angled triangle for details).

The quadrilateral $A D E C$ therefore splits into three isosceles triangles, as in the following diagram.

Now, the interior angles of a quadrilateral add up to $360^\circ$, so we have:

As triangle $A D C$ is half of a square, angles $A \hat{D} O$ and $O \hat{A} D$ are both $45^\circ$. Together with the fact that triangles $O E C$ and $O D E$ are isosceles, this means that we have:

Hence angle $D \hat{E} C = 135^\circ$.

From this the desired angle can be easily calculated, since angle $D \hat{E} H = D \hat{E} C - 90^\circ = 45^\circ$ and angle $G \hat{H} E$ is a right-angle. Therefore angle $H \hat{G} E$ is $45^\circ$ as angles in a triangle add up to $180^\circ$ and so angle $F \hat{G} E = 135^\circ$ as angles on a straight line also add up to $180^\circ$.

Solution by Circles

The triangle $A C D$ is a right-angled triangle with hypotenuse $A C$. Mark a point $O$ at the midpoint of $A C$. Then the circle centre $O$ with diameter $A C$ also goes through $D$.

The triangle $A C E$ is also a right-angled triangle with hypotenuse $A C$. So the same circle also goes through $E$.

There are now two routes to get angle $D \hat{E} A$.

One route notes that $A C E D$ is a cyclic quadrilateral and so $D \hat{E} C + C \hat{A} D = 180^\circ$. Since $C \hat{A} D = 45^\circ$, this shows that $D \hat{E} E = 135^\circ$ and so $D \hat{E} A = 45^\circ$.

The other route uses the chord $D A$ with angles in the same segment to see that $D \hat{E} A = D \hat{C} A = 45^\circ$.

To get angle $F \hat{G} D$, we then note that $G \hat{H} D = 90^\circ$ so $H \hat{G} D = 45^\circ$ and so $F \hat{G} D = 135^\circ$.