# Solution to the Two Circles Inside an Equilateral Triangle Puzzle +-- {.image} [[TwoCirclesInsideanEquilateralTriangle.png:pic]] > The triangle is equilateral. What’s the angle? =-- ## Solution by [[Isosceles Triangle]] and [[Angles in a Quadrilateral]] +-- {.image} [[TwoCirclesInsideanEquilateralTriangleLabelled.png:pic]] =-- 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]] +-- {.image} [[TwoCirclesInsideanEquilateralTriangleSpecial.png:pic]] =-- 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 $150^\circ$.