# Solution to the [[Two Equilateral Triangles III]] Puzzle +-- {.image} [[TwoEquilateralTrianglesIII.jpeg:pic]] > Two equilateral triangles. What’s the angle? =-- ## Solution by Properties of [[Equilateral Triangles]], [[Vertically Opposite Angles]], [[Angles at a Point on a Straight Line]], and [[Congruent]] Triangles +-- {.image} [[TwoEquilateralTrianglesIIILabelled.jpeg:pic]] =-- 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. +-- {.image} [[TwoEquilateralTrianglesIIIInvarianceA.jpeg:pic]] =-- In this version of the diagram, the two equilateral triangles are the same size and the requested angle is then $180^\circ - 60^\circ - 60^\circ = 60^\circ$. +-- {.image} [[TwoEquilateralTrianglesIIIInvarianceB.jpeg:pic]] =-- 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 $60^\circ$.