# Solution to the Isosceles and Equilateral Triangles Puzzle +-- {.image} [[IsoscelesandEquilateralTriangles.png:pic]] > A blue equilateral triangle overlaps a pink isosceles triangle. What’s the angle? =-- ## Solution by Direct Calculation Consider the labelled diagram. +-- {.image} [[IsoscelesandEquilateralTrianglesLabelled.png:pic]] =-- Let $x = A\hat{B} D$ and $y = D\hat{B} C$. Then $x + y = 60^\circ$ as the [[interior angle]] in an [[equilateral triangle]]. As triangle $A B D$ is [[isosceles]], angle $A \hat{D} B$ is half of $180^\circ - x$. Also, as it is isosceles then $D B = A B$, then as triangle $A B C$ is [[equilateral]], $D B = C B$ so triangle $D B C$ is also isosceles. By the same reasoning, then, angle $B \hat{D} C$ is half of $180^\circ - y$. Putting these together, angle $A \hat{D} C$ is: $$ \frac{180^\circ - x + 180^\circ - y}{2} = 180^\circ - \frac{x + y}{2} = 180^\circ - 30^\circ = 150^\circ $$ ## Solution using Circles Since triangle $A B D$ is isosceles and $A B C$ is equilateral, drawing a circle centred on $B$ with radius $A B$ passes through both $C$ and $D$. +-- {.image} [[IsoscelesandEquilateralTrianglesWithCircle.png:pic]] =-- Using the [[angle on the circumference|angle on the circumference is half the angle at the centre]] then angle $A \hat{D} C$ is half of angle $A \hat{B} C$, where the latter is the reflex angle. Hence: $$ A \hat{D} C = \frac{360^\circ - 60^\circ}{2} = 150^\circ $$