Solution with complex numbers Let $\frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB}=k$ $d=\frac{b+k \cdot c}{k+1}$ $e=\frac{c+ k \cdot a}{k+1}$ $f=\frac{a+k \cdot b}{k+1}$ Let the origin be $O$ - the circumcenter of $\triangle{ABC}$ and $\triangle DEF$ So: $|a|=|b|=|c| \implies |a|^2=|b|^2=|c|^2 \implies a \cdot \overline a = b \cdot \overline b = c \cdot \overline c$ And: $|d|=|e|=|f| \implies |d|^2=|e|^2=|f|^2 \implies d \cdot \overline{d}=e \cdot \overline{e}=f \cdot \overline{f}$ $\implies (b + k \cdot c)(\overline b + k \cdot \overline c)=(c + k \cdot a)(\overline c + k \cdot \overline a)$ $\implies b \cdot \overline b + k^2 \cdot c \cdot \overline c + k \cdot c \cdot \overline b + k \cdot \overline c \cdot b = c \cdot \overline c + k^2 \cdot a \cdot \overline a + k \cdot a \cdot \overline c + k \cdot \overline a \cdot c$ $\implies c \cdot \overline b + \overline c \cdot b = a \cdot \overline c + \overline a \cdot c$ Similar, we have $ c \cdot \overline b + \overline c \cdot b = a \cdot \overline c + \overline a \cdot c = b \cdot \overline a + \overline b \cdot a$ I want $|a-b|=|b-c|=|c-a|$ $|a-b|=|c-a| \iff |a-b|^2=|c-a|^2 \iff (a-b)( \overline a - \overline b) = (c-a)( \overline c -\overline a)$ $\iff a \cdot \overline a + b \cdot \overline b - b \cdot \overline a - a \cdot \overline b = c \cdot \overline c + a \cdot \overline a - c \cdot \overline a - a \cdot \overline c \iff 0=0$, true Similar, we have $|a-b|=|b-c|=|c-a| \iff AB=BC=CA$, so $ABC$ is equilateral $\blacksquare$