Notes
two equilateral triangles ii solution

Solution to the Two Equilateral Triangles II Puzzle

Solution by Transformations and Lengths and Angles in an Equilateral Triangle

Consider the diagram labelled as above.

Triangle $A C F$ is similar to triangle $A B E$ by a rotation of $30^\circ$ anticlockwise about $A$ with a scaling with scale factor $\frac{\sqrt{3}}{2}$. This comes from the lengths and angles in an equilateral triangle. In particular, $C F$ is at angle $30^\circ$ to $B E$.

The requested angle is the obtuse angle $B \hat{H} F$, which is then $180^\circ - 30^\circ = 150^\circ$.

Solution by Invariance Principle

The relative sizes of the triangles is not fixed, so they can be drawn at different ratios.

In this first version the two triangles are the same size, so line segments $A C$ and $A F$ are the same length and thus triangle $A C F$ is equilateral triangle. Hence angle $F \hat{C} A = 60^\circ$, so since angle $A \hat{C} B = 90^\circ$, the requested angle is $90^\circ + 60^\circ = 150^\circ$.

In the second version the right-hand triangle is shrunk to a point, so $A$, $E$, $F$, $G$, and $H$ coincide. To get the requested angle we extend $B E$ and $C F$, then as $B \hat{A} C = 30^\circ$, the requested angle is $150^\circ$.