Given an equilateral triangle ABC and P is any point inside the triangle such that PA² =PB² +PC² ,then angle BPC is

Let's position the triangle so that B( 0,1/2 ) , C ( 0,-1/2)

and A( sqrt(3) /2, 0 )

Here the scale of the triangle is not important.

Let P(x,y)

then the condition PA^2 = PB^2 +PC^2 can be written

x^2 +y^2 + sqrt(3)x - 1/4 = 0

In other words, the locus of P is this circle .
Actually the locus is only an arc of this circle because P

must lie within the triangle.
This circle will cut the Ox axis

at A1 with OA1 = 1 - (sqrt(3) /2 ) ( concentrate on this one ) { let y=0 into

circle equation }

and A2 with the coordinates of A2 is [ ( -sqrt(3)/2 - 1 ) ,0 ]

let alpha = angle < B A1 O

tan (alpha) =3.732

==> alpha = 1.3089 rad = 75 deg

then angle B A1 C = 2 (75) =150 deg

Also angle BPC = 150 deg