Notes
two circles inside an equilateral triangle solution

Solution to the Two Circles Inside an Equilateral Triangle Puzzle

Solution by Isosceles Triangle and Angles in a Quadrilateral

In the above diagram, the line segment $D F$ is tangent to the two circles at $H$. The lengths of $F G$ and $F H$ are the same, so triangle $F G H$ is isosceles and so angles $H \hat{G} F$ and $F \hat{H} G$ are equal. Extending $H F$ and $E C$ to the point where they meet creates another isosceles triangle to show that angles $E \hat{H} F$ and $C \hat{E} H$ are also equal. The sum of the four angles $C \hat{E} H$, $E \hat{H} F$, $F \hat{H} G$, and $H \hat{G} F$ is then twice the marked angle. As the outer triangle is equilateral, angle $A \hat{C} B$ is $60^\circ$. As the angles in a quadrilateral sum to $360^\circ$, this means that angles $C \hat{E} H$, $E \hat{H} G$, and $H \hat{G} C$ sum to $300^\circ$ meaning that the marked angle is $150^\circ$.

Solution by Invariance Principle

In the above diagram, the two circles are the same size as each other. The red line is then vertical and goes through point $C$. Angles $G \hat{C} H$ and $H \hat{C} E$ are both $30^\circ$ and triangles $H C G$ and $E C H$ are isosceles (and congruent) with base angles half of $180^\circ - 30^\circ = 150^\circ$. So angle $E \hat{H} G$ is $150^\circ$.