Point in an equilateral $\triangle$

Let the coordinates of $A (0, \sqrt 3)$ , $B(-1,0)$ , $C(0,1)$ , and $P(x,y)$ . Then

$\begin{aligned} AP^2 & = BP^2+CP^2 \\ (x-0)^2+(y-\sqrt 3)^2 & = (x+1)^2+(y-0)^2+(x-1)^2+(y-0)^2 \\ x^2 + y^2-2\sqrt 3y +3 & = 2x^2 + 2 + 2y^2 \\ x^2 + y^2 + 2\sqrt 3y & = 1 \\ x^2 + y^2 - 1 & = - 2\sqrt 3y \quad \color{#3D99F6}...(*) \end{aligned}$

Let the foot of altitude from $P$ to $BC$ be $Q$ , then

$\begin{aligned} \angle BPC & = \angle BPQ + \angle CPQ \\ \tan (\angle BPC) & = \tan (\angle BPQ + \angle CPQ) \\ & = \frac {{\color{#3D99F6}\tan (\angle BPQ)} + \color{#D61F06}\tan(\angle CPQ)}{1 - {\color{#3D99F6}\tan (\angle BPQ)} \color{#D61F06}\tan(\angle CPQ)} \\ & = \frac {{\color{#3D99F6}\frac {1+x}y} + \color{#D61F06}\frac {1-x}y}{1 - {\color{#3D99F6}\frac {1+x}y} \times \color{#D61F06}\frac {1-x}y} \\ & = \frac {2y}{\color{#3D99F6} x^2+y^2-1} & \small \color{#3D99F6} (1): \quad x^2 + y^2 - 1 = - 2\sqrt 3y \\ & = \frac {2y}{\color{#3D99F6} - 2\sqrt 3y} = - \frac 1{\sqrt 3} \\ \implies \angle BPC & = \boxed{150}^\circ \end{aligned}$