Notes
two squares and an equilateral triangle ii solution

Solution to the Two Squares and an Equilateral Triangle II Puzzle

Solution by Pythagoras' Theorem, Lengths and Angles in an Equilateral Triangle, and Angles in an Isosceles Triangle

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:

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

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 $75^\circ$.