Notes
two squares and an equilateral triangle ii solution

Solution to the Two Squares and an Equilateral Triangle II Puzzle

Two Squares and an Equilateral Triangle II

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

Two squares and an equilateral triangle annotated

Label the diagram as above.

Let ABA B have length 11, then from the lengths in an equilateral triangle, FBF B has length 3\sqrt{3} so the side length of the squares is 32\frac{\sqrt{3}}{2}. Angles ED^FE \hat{D} F and BF^DB \hat{F} D are alternate angles so ED^FE \hat{D} F is 30 30^\circ. This means that triangle FEDF E D is half an equilateral triangle, so EDE D has length 3\sqrt{3} times that of FEF E, so has length 32\frac{3}{2}. Applying Pythagoras' Theorem to triangle FEDF E D shows that the length of FDF D is the square root of:

(32) 2+(32) 2=34+94=124=3 \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{3}{2}\right)^2 = \frac{3}{4} + \frac{9}{4} = \frac{12}{4} = 3

Hence FDF D has length 3\sqrt{3} which is the same as the length of FBF B. Therefore, triangle FBDF B D is isosceles. Since angle BF^DB \hat{F} D is 30 30^\circ, this means that angle FD^BF \hat{D} B is 75 75^\circ.

Solution by Lengths and Angles in an Equilateral Triangle, and Angles in an Isosceles Triangle

Two squares and an equilateral triangle annotated b

Extend the diagram by adding two more squares as above, and label the diagram as shown.

By symmetry, triangle FDGF D G is isosceles. Then since angles FC^BF \hat{C} B and CF^GC \hat{F} G are corresponding angles they are the same, so angle DF^GD \hat{F} G is 60 60^\circ. This means that triangle FDGF D G is actually equilateral and so FDF D has the same length as FGF G, which has the same length as FBF B.

Hence triangle FDBF D B is isosceles and since angle BF^DB \hat{F} D is 30 30^\circ, angle FD^BF \hat{D} B is 75 75^\circ.