Pythagorean point in Equilateral triangle

construct an equilateral triangle PDC such that D is a point outside triangle ABC somewhere below BC. now PD = PC = CD.let angle BCP = x then BCD = 60 - x and PCA = 60 - x . now in triangle BDC and APC BC = AC , BCD = PCA = 60 - x and PC = PD (as constructed). so they are congruent. now AP = BD by cpct PC = PD therefore ${ BD }^{ 2 }\quad =\quad { { { PB } } }^{ 2\quad }+\quad { PD }^{ 2 }\\ we\quad find\quad pythagorous\quad theorem\quad true\quad in\quad this\quad triangle\\ thus\quad angle\quad BPD\quad =\quad 90\quad \\ and\quad BPC\quad =\quad BPD\quad +\quad DPC\quad \\ BPC\quad =\quad 90\quad +\quad 60\quad =\quad 150$

Let be $P'$ a point outside of $\triangle ABC$ so that $P'C=PC$ and $P'B=PA$ . $\Rightarrow \angle P'CP=60°$ and as $P'C=PC$ , $\angle CP'P=\angle CPP'=60° \therefore \triangle PP'C$ is equilateral. Also, $\triangle P'PB$ is right so $\angle P'PB=90°$

$\angle BPC=\angle P'PB+\angle CP'P=90°+60°=\boxed{150°}$