Notes
three squares viii solution

Solution to the Three Squares VIII Puzzle

Solution by Similar Triangles and Gradients of Perpendicular Lines

In the above diagram, line segment $E C$ is vertical through the intersection point, $D$.

By similarity, the lengths of $D C$ and $C A$ are in the ratio $1 : 2$ and the lengths of $D E$ and $E F$ are in the ration $1 : 3$. Therefore, the lengths of $C D$ and $D E$ are in the ratio $3 : 2$.

The length of $C D$ is then $\frac{3}{5}$ of the side length of a square, so $A C$ has length $2 \times \frac{3}{5} = \frac{6}{5}$ of the side length of a square. Then $B C$ has length $\frac{6}{5} - 1 = \frac{1}{5}$ of the side length of a square.

The gradient of $B D$ is therefore:

The gradient of $F D$ is $-\frac{1}{3}$. Therefore $F D$ and $D B$ are perpendicular and so the angle is $90^\circ$.