<s>[INST] Clarification in proof of perpendicular bisectors meeting at a point Context:

My crude drawing from Paint to illustrate triangle OAB:


My working:
$\begin{align*}
(z-\frac{1}{2}(x+y)) \cdot (y-x) &= z\cdot(y-x) + \frac{1}{2}(-x-y) \cdot (y-x)\\
&= z\cdot y -z\cdot x + \frac{1}{2}(\lVert x\rVert^{2} - \lVert y\rVert^{2})\\
&= (\frac{1}{2}y+b\rho(y))\cdot y - (\frac{1}{2}x+c\rho(x))\cdot x+ \frac{1}{2}(\lVert x\rVert^{2} - \lVert y\rVert^{2})\\
&= \frac{1}{2}y\cdot y - \frac{1}{2}x\cdot x+ \frac{1}{2}(\lVert x\rVert^{2} - \lVert y\rVert^{2}) \\
&= \frac{1}{2}\lVert y\rVert^{2} - \frac{1}{2}\lVert y\rVert^{2} - \frac{1}{2}\lVert x\rVert^{2} + \frac{1}{2}\lVert x\rVert^{2}\\
&= 0
\end{align*}$
However, I can't see how this helps me "show that $z$ lies on the perpendicular bisector of $\vec{AB}$ " as we have only shown $(z-\frac{1}{2}(x+y))$ (whatever this point is) is perpendicular to $(y-x)=\vec{AB}$.
Cheers!
 [/INST] Let $K$ be the point of intersection of the perpendicular bisectors of the sides $OA$ and $OB.$ Let $\overrightarrow {OK} = z.$ Let $M$ be the midpoint of the side $AB.$ Then observe that if we can show that $\overrightarrow {MK} \perp \overrightarrow {AB}$ we are through. So we need only to show that $$\overrightarrow {MK} \cdot \overrightarrow {AB} = 0.$$ Now what is $\overrightarrow {MK}$?
Observe that $$\overrightarrow {MK} = \overrightarrow {OK} - \overrightarrow {OM}.$$ Now observe that $$\overrightarrow {AM} = \frac {1} {2} (y-x)\ \ \text {and}\ \overrightarrow {OM} = \overrightarrow {OA} + \overrightarrow {AM} = x+ \frac {1} {2} (y-x) = \frac {1} {2} (x+y).$$ Also $\overrightarrow {OK} = z.$ So we have $$\overrightarrow {MK} = z - \frac {1} {2} (x+y).$$ Again $\overrightarrow {AB} = y-x.$ So we need only to show that $$\left (z - \frac {1} {2} (x+y) \right ) \cdot (y-x) = 0$$ as required.
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