Finding angle in an equilateral triangular pyramid

Question

Given an equilateral triangular pyramid (refer the below diagram) $\Delta ABC$ & $P$ is any point inside the triangle such that ${PA}^{2}={PB}^{2}+{PC}^{2}$, then $\angle BPC$ is -

I am unable to think of how to do this question

Answer 1

Here's a complete solution. (I think)

Define an equilateral triangle on the coordinate system as follows:

$A=(0,{\sqrt3\over 2}),B=(0.5,0),C=(-0.5,0),P=(x,y)$

By the requirement of $P$$$y^2 +(x+0.5)^2+y^2+(x-0.5)^2=x^2+(y-{\sqrt3\over 2})^2$$$$\implies \left(y+{\sqrt3\over 2}\right)^2+x^2=1 $$

$\therefore$ The locus of $P$ is the circle with center $\left( 0,-{\sqrt3\over 2}\right)$ and radius $1$.

Let the center be $O$ So $\angle BPC=1/2(\angle BOC)=150^{\circ}$ ( Knowing the coordinates of $ B;O;C$) Problem SOLVED!!

Answer 2

Hint: rotate the triangle $60^{\circ}$ clockwise around $B$, so that $A$ is rotated onto $C$, and let $P'$ be the image of $P$ under this rotation. Can you show the following statements:

1. $\angle PBP' = 60^{\circ}$
2. $\triangle PBP'$ is equilateral
3. $PP' = PB$
4. $P'C = PA$
5. $\triangle P'PC$ is a right triangle, with a right angle at $P$
6. $\angle BPC = \angle BPP' + \angle P'PC$

If you can show these statements, then the answer should follow.