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
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$\begingroup$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!!
$\endgroup$ 9 $\begingroup$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:
- $\angle PBP' = 60^{\circ}$
- $\triangle PBP'$ is equilateral
- $PP' = PB$
- $P'C = PA$
- $\triangle P'PC$ is a right triangle, with a right angle at $P$
- $\angle BPC = \angle BPP' + \angle P'PC$
If you can show these statements, then the answer should follow.
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