Let $ABC$ be an original equilateral triangle with center at point $O$. Then, answer is all points $P$, satysfying at least one of three conditions $\angle APB=120^{\circ}$ or $\angle BPC=120^{\circ}$ or $\angle CPA=120^{\circ}$.

Denote distance from $P$ to line $\ell$ as $\rho(P,\ell)$. Suppose that $\rho^2(P,AB)=\rho(P,AC)\rho(P,BC)$. We will prove that $\angle APB = 120^{\circ}$. Since $AB=AC=BC$ we conclude that $S^2_{APB}=S_{APC}S_{BPC}$. Let $A_1=AP~\cap~BC$ and $B_1=BP~\cap~AC$. Observe that $\frac{BA_1}{CA_1}=\frac{S_{APB}}{S_{APC}}$ and $\frac{AB_1}{CB_1}=\frac{S_{APB}}{S_{BPC}}$. Hence, $\frac{BA_1}{CA_1}=\frac{CB_1}{AB_1}$. From this we get $BA_1=CB_1$ and $A_1C=B_1A$. Therefore $\angle APB=120^{\circ}$. For such points equality $\rho^2(P,AB)=\rho(P,AC)\rho(P,BC)$ can easy proved.