Notes
two equilateral triangles iii solution

Solution to the Two Equilateral Triangles III Puzzle

Solution by Properties of Equilateral Triangles, Vertically Opposite Angles, Angles at a Point on a Straight Line, and Congruent Triangles

Consider the diagram as labelled above. In this diagram, line segment $G D$ is the continuation of $A G$.

Since vertically opposite angles are equal, angle $C \hat{G} D$ is the same as angle $F \hat{G} A$, hence is $60^\circ$. Therefore, two of the angles in triangle $G C D$ are $60^\circ$ and so triangle $G C D$ is equilateral.

Therefore, line segments $C G$ and $C D$ have the same length, so so also do line segments $G F$ and $D E$. Angles $D \hat{G} F$ and $E \hat{D} G$ are both $120^\circ$, since angles at a point on a straight line add up to $180^\circ$, so triangles $F G D$ and $E D G$ are congruent. In particular, angle $D \hat{F} G$ is the same as angle $D \hat{E} G$.

Similarly, line segments $F G$ and $A G$ have the same length, as do $G D$ and $G C$. Then also angles $A \hat{G} C$ and $D \hat{G} F$ are equal. So triangles $F G D$ and $A G C$ are also congruent. In particular, angles $G \hat{A} C$ and $D \hat{F} G$ are equal.

So angle $G \hat{A} C$ is equal to angle $G \hat{E} C$. Moreover, angles $A \hat{G} B$ and $E \hat{G} D$ are equal as they are vertically opposite. Therefore, angles $G \hat{B} A$ and $E \hat{D} G$ are also equal, but $E \hat{D} G$ is equal to $120^\circ$. So angle $C \hat{B} G$ is equal to $180^\circ - 120^\circ = 60^\circ$ since angles at a point on a straight line add up to $180^\circ$.

Solution by Invariance Principle

The relative sizes of the equilateral triangles can vary, leading to the following two configurations.

In this version of the diagram, the two equilateral triangles are the same size and the requested angle is then $180^\circ - 60^\circ - 60^\circ = 60^\circ$.

In this version, one triangle has shrunk down to a point and the requested angle is the interior angle of an equilateral triangle, so is $60^\circ$.