# Solution to the [[Three Squares VIII]] Puzzle +-- {.image} [[ThreeSquaresVIII.jpeg:pic]] > Three squares. What’s the angle? =-- ## Solution by [[Similar Triangles]] and [[Gradients of Perpendicular Lines]] +-- {.image} [[ThreeSquaresVIIILabelled.jpeg:pic]] =-- 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: $$ \frac{ \phantom{x}\frac{3}{5}\phantom{x} }{ \frac{1}{5} } = 3 $$ The gradient of $F D$ is $-\frac{1}{3}$. Therefore $F D$ and $D B$ are [[perpendicular]] and so the angle is $90^\circ$.