Question 1166456
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Assume P,Q,and R are oriented in an arbitrary way WRT a set of coordinate axes (but the points obey the constraints given by the problem statement).
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Now rotate and translate a set of coordinate axes (WLOG) such that Q is at the origin {{{0+0i}}}, P is at {{{a[1]+b[1]i}}} and R is at {{{a[3] + 0i}}}:
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{{{drawing(300,300, -2, 8, -2, 8, grid(0),
           line(0,0,3,3),
           line(3,3,6,0),
           locate(0.1,-0.2,Q), 
           locate(3,3.8,P (a[1]+b[1]i)),
           locate(6.1,1.1,R (a[3]))
) }}}
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Then you get:
LHS = {{{2z[1]^2+z[2]^2+z[3]^2 = 2z[1]^2 + a[3]^2 }}}

RHS = {{{2z[1](z[2]+z[3]) = 2(a[1]+b[1]i)(a[3]) = 2a[1]a[3]+2a[3]b[1]i }}}


Apply Pythagorean Theorem {{{ abs(PQ)^2 + abs(PR)^2 = abs(QR)^2 }}}:<br>
{{{ (a[1]-a[3])^2+b[1]^2+a[1]^2+b[1]^2 = a[3]^2 }}} 
...
   {{{2(a[1]^2 + b[1]^2) = 2a[1]a[3] }}}
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Note that {{{a[3] = 2a[1]}}} so the LHS becomes:  {{{2z[1]^2 + a[3]^2 = 2z[1]^2 + 4a[1]^2 }}} <br>
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and using the Pythagorean result, the RHS becomes:  {{{ 2a[1]a[3]+2a[3]b[1]i = 2a[1]^2+2b[1]^2 + 4a[1]b[1]i = 2z[1]^2 + 4b[1]^2 }}}<br>

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Drop a perpendicular from P to the x-axis, the height of P from the x-axis and horizontal distance from origin are the same, these are also {{{b[1] }}} and {{{a[1]}}} respectively.  Thus, we have shown {{{a[1] = b[1]}}}.
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Siince {{{a[1] = b[1]}}} the proof is complete.

Undoing the translation and rotation of the coordinate axes back to whatever the 'original' position was changes nothing (but the algebra is more complicated in that orientation).