Equilateral triangles $ABX$ and $CAY$ are described on sides $AB$ and $AC$ of a $\triangle ABC$ externally to $\triangle ABC$. Prove that $CX = BY$.

Question

Equilateral triangles $ABX$ and $CAY$ are described on sides $AB$ and $AC$ of a $\triangle ABC$ externally to $\triangle ABC$. Prove that $CX = BY$.

I constructed the following figure for it.

I am not able to proceed any further. How could I do this?

Answer 1

Hint: $\triangle ACX$ and $\triangle BAY$ are equal to each other: $$|AC|=|AY|, |AX|=|AB|$$and $Â$ in $\triangle BAY $ is equal to $Â$ in $\triangle ACX$ because $\triangle ABX$ and $\triangle ACY$ are equilateral triangles.

Answer 2

Hint:

• What would you get if you were to rotate point $X$ around $A$ counterclockwise by $\frac{\pi}{3}$?
• What would you get if you were to rotate point $C$ around $A$ counterclockwise by $\frac{\pi}{3}$?
• What would you get if you were to rotate $\triangle ACX$ around $A$ counterclockwise by $\frac{\pi}{3}$?

I hope this helps $\ddot\smile$

Answer 3

Let $A$ be the origin and $z_2, z_3$ complex numbers representing $B, C$ respectively. $Y$ is $z_3e^{ipi/3}$ and $X$ is $z_2e^{-i \pi/3}$. \begin{align*} CX = |z_3 - z_2e^{-i \pi/3}| = |e^{-i\pi/3}(z_3e^{i\pi/3} - z_2)|=|z_3e^{i\pi/3} - z_2| = BY \end{align*}