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Context:

enter image description here

My crude drawing from Paint to illustrate triangle OAB:

enter image description here


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!

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  • $\begingroup$ Perpendicular bisectors are not necessarily passing through the vertices of a triangle. Draw the diagram clearly. What you have drawn holds only if the triangle is equilateral. In which case it is easy to see that $z$ is perpendicular to $y-x$ as $\|x\| = \|y\|.$ $\endgroup$ – Dbchatto67 Apr 22 at 10:41
  • $\begingroup$ @Dbchatto67 Hi I do know that which is why I deliberately drew the angle bisector $\frac{1}{2}(x+y)$ so that it doesn't meet $z$. $\endgroup$ – Darius Apr 22 at 10:46
  • $\begingroup$ There is no reason to consider the vector $z$ in this picture. What you have to show? You take perpendicular bisectors of the sides $OA$ and $OB.$ Let they meet at $K.$ Where $\overrightarrow {OK} = z.$ Now you take the vector joining $K$ and the midpoint $M$ of $AB.$ If you can show that $\overrightarrow {KM} \perp \overrightarrow {AB}$ you are done. Right? What is the vector $\overrightarrow {KM}$? $\endgroup$ – Dbchatto67 Apr 22 at 10:58
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    $\begingroup$ Have you noticed that $$\overrightarrow {MK} = z - \frac {1} {2} (x+y)?$$ $\endgroup$ – Dbchatto67 Apr 22 at 11:16
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    $\begingroup$ Because we have $$\overrightarrow {MK} = \overrightarrow {OK} - \overrightarrow {OM}.$$ 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 $$\overrightarrow {MK} = z - \frac {1} {2} (x+y)$$ as required. $\endgroup$ – Dbchatto67 Apr 22 at 11:24
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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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