# Solution to the [[Two Equilateral Triangles II]] Puzzle +-- {.image} [[TwoEquilateralTrianglesII.jpeg:pic]] > Both triangles are equilateral. What’s the angle? =-- ## Solution by [[Transformations]] and Lengths and Angles in an [[Equilateral Triangle]] +-- {.image} [[TwoEquilateralTrianglesIIAnnotated.jpeg:pic]] =-- Consider the diagram labelled as above. Triangle $A C F$ is [[similar]] to triangle $A B E$ by a [[rotation]] of $30^\circ$ anticlockwise about $A$ with a [[scaling]] with scale factor $\frac{\sqrt{3}}{2}$. This comes from the lengths and angles in an [[equilateral triangle]]. In particular, $C F$ is at angle $30^\circ$ to $B E$. The requested angle is the obtuse angle $B \hat{H} F$, which is then $180^\circ - 30^\circ = 150^\circ$. ## Solution by [[Invariance Principle]] The relative sizes of the triangles is not fixed, so they can be drawn at different ratios. +-- {.image} [[TwoEquilateralTrianglesIIInvariantA.png:pic]] =-- In this first version the two triangles are the same size, so line segments $A C$ and $A F$ are the same length and thus triangle $A C F$ is [[equilateral|equilateral triangle]]. Hence angle $F \hat{C} A = 60^\circ$, so since angle $A \hat{C} B = 90^\circ$, the requested angle is $90^\circ + 60^\circ = 150^\circ$. +-- {.image} [[TwoEquilateralTrianglesIIInvariantB.png:pic]] =-- In the second version the right-hand triangle is shrunk to a point, so $A$, $E$, $F$, $G$, and $H$ coincide. To get the requested angle we extend $B E$ and $C F$, then as $B \hat{A} C = 30^\circ$, the requested angle is $150^\circ$.