In the equilateral triangle $ABC$, points $D,E$ move on extensions of $AB$ and $AC$ so that $BD.CE=BC^2$.Call intesection of $CD,BE$ as $K$.Find the geometric path that $K$ traverses.
I tried to use notion of power of a point in order to solve this but couldn't find a candidate circle for which the power of a given point becomes $BC^2$.
Hint: $\frac{BD}{BC} = \frac{BC}{CE}$ and $\widehat{CBD} = \widehat{BCE}$ so $\triangle BCD$ and $\triangle CEB$ are similar, thus $\widehat{CDB} = \widehat{EBC}$.
Then $\widehat{BKC} = \pi - \widehat{CBE} - \widehat{BCD} = \pi - \widehat{CDB} - \widehat{BCD}$. But from $\triangle BDC$ $\;\;\widehat{CDB} + \widehat{BCD} = ...$
