# Solution to the [[Three Squares VI]] Puzzle +-- {.image} [[ThreeSquaresVI.jpeg:pic]] > The blue and green shaded regions created by these three squares have the same area. What’s the angle? =-- ## Solution by [[Pythagoras' Theorem]] and [[Angles in a Quadrilateral]] +-- {.image} [[ThreeSquaresVILabelled.jpeg:pic]] =-- Let the side lengths of the squares be $a$, $b$, and $c$ in increasing order. Then $E B$ has length $a + b - c$ so the green region has area $a b + b^2 - c b$. The line segment $F C$ has length $c - a$, so the blue region has area: $$ c^2 - b(c - a) - a^2 = c^2 - b c + b a - a^2 $$ These areas are the same, so: $$ \begin{aligned} c^2 - b c + b a - a^2 &= a b + b^2 - c b \\ c^2 - a^2 &= b^2 \\ c^2 &= a^2 + b^2 \end{aligned} $$ This means that the lengths $a$, $b$, and $c$ form the sides of a [[right-angled triangle]], by the converse to [[Pythagoras' theorem]]. Triangle $A B C$ is a right-angled triangle with sides $a$ and $b$, so its hypotenuse, $A B$, has length $c$. This means that triangles $A D B$ and $F A B$ are both [[isosceles]] with angles $A \hat{D} B$ and $B \hat{A} D$ equal to each other, and $B \hat{F} A$ and $F \hat{A} B$ also equal to each other. Then as the [[angles in a quadrilateral]] add up to $360^\circ$: $$ \begin{aligned} 360^\circ &= B \hat{F} A + F \hat{A} B + B \hat{A} D + A \hat{D} B + D \hat{B} F \\ &= 2 F \hat{A} B + 2 B \hat{A} D + 90^\circ \\ 270^\circ &= 2 F \hat{A} D \\ 135^\circ &= F \hat{A} D \end{aligned} $$ Hence the angle is $135^\circ$.