[[!redirects two squares and an equilateral triangle ii solutions]] # Solution to the [[Two Squares and an Equilateral Triangle II]] Puzzle +-- {.image} [[TwoSquaresandanEquilateralTriangleII.jpeg:pic]] > Two squares and an equilateral triangle. What’s the angle? =-- ## Solution by [[Pythagoras' Theorem]], Lengths and Angles in an [[Equilateral Triangle]], and Angles in an [[Isosceles Triangle]] +-- {.image} [[TwoSquaresandanEquilateralTriangleIIAnnotated.jpeg:pic]] =-- Label the diagram as above. Let $A B$ have length $1$, then from the [[lengths in an equilateral triangle]], $F B$ has length $\sqrt{3}$ so the side length of the squares is $\frac{\sqrt{3}}{2}$. Angles $E \hat{D} F$ and $B \hat{F} D$ are [[alternate angles]] so $E \hat{D} F$ is $30^\circ$. This means that triangle $F E D$ is half an [[equilateral triangle]], so $E D$ has length $\sqrt{3}$ times that of $F E$, so has length $\frac{3}{2}$. Applying [[Pythagoras' Theorem]] to triangle $F E D$ shows that the length of $F D$ is the square root of: $$ \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{3}{2}\right)^2 = \frac{3}{4} + \frac{9}{4} = \frac{12}{4} = 3 $$ Hence $F D$ has length $\sqrt{3}$ which is the same as the length of $F B$. Therefore, triangle $F B D$ is [[isosceles]]. Since angle $B \hat{F} D$ is $30^\circ$, this means that angle $F \hat{D} B$ is $75^\circ$. ## Solution by Lengths and Angles in an [[Equilateral Triangle]], and Angles in an [[Isosceles Triangle]] +-- {.image} [[TwoSquaresandanEquiTriangleIIAnnotatedB.jpeg:pic]] =-- Extend the diagram by adding two more squares as above, and label the diagram as shown. By [[symmetry]], triangle $F D G$ is [[isosceles]]. Then since angles $F \hat{C} B$ and $C \hat{F} G$ are [[corresponding angles]] they are the same, so angle $D \hat{F} G$ is $60^\circ$. This means that triangle $F D G$ is actually [[equilateral]] and so $F D$ has the same length as $F G$, which has the same length as $F B$. Hence triangle $F D B$ is [[isosceles]] and since angle $B \hat{F} D$ is $30^\circ$, angle $F \hat{D} B$ is $75^\circ$.