Notes
a triangle inside a hexagon solution

Solution to the A Triangle Inside a Hexagon Puzzle

Solution by Congruent Triangles, Angles in a Regular Hexagon, and Angles in an Equilateral Triangle

In the above diagram, $O$ is the centre of the hexagon. Although $O H F$ looks like a straight line - and will turn out to be a straight line - that is not a given from the diagram.

Consider triangles $A G O$ and $A H F$. Side lengths $A G$ and $A H$ have the same length, as do $A O$ and $A F$.

Triangles $B A O$, $G A H$, and $O A F$ are all equilateral triangles, so angles $G \hat{A} H$ and $O \hat{A} F$ are both $60^\circ$. Therefore, angles $G \hat{A} O$ and $H \hat{A} F$ are equal.

This establishes that triangles $A G O$ and $A H F$ are congruent. Hence angle $A \hat{F} H$ is the same as angle $A \hat{O} G$, which is $60^\circ$. Since the interior angle in a regular hexagon is $120^\circ$, angle $H \hat{F} E$ is $60^\circ$.

Solution by Rotation

This is an application of the weak invariance principle. The point $G$ is not fixed on the line $B E$ and as $G$ moves along this line, so then $H$ moves as well.

To identify the possible positions of $H$, consider a rotation of angle $60^\circ$ anticlockwise about point $A$. This takes point $B$ to $O$, $G$ to $H$, and $O$ to $F$. Therefore, since $B G O$ is a straight line, so also is $O H F$. Hence, $H$ always lies on the line $O F$.

This establishes that angle $H \hat{F} E$ is the same as angle $O \hat{F} E$, which is $60^\circ$.

Solution by Invariance Principle

Since $G$ is free to move along the line segment $B E$, there are configurations where the angle is straightforward to see. In the above, $G$ coincides with $B$ which brings $H$ to $O$ and hence angle $H \hat{F} E$ is $60^\circ$.