# 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!

• 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 $\|x\| = \|y\|.$ – Dbchatto67 Apr 22 at 10:41
• @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 $z$. – Darius Apr 22 at 10:46
• 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 $\overrightarrow {KM}$? – Dbchatto67 Apr 22 at 10:58
• Have you noticed that $$\overrightarrow {MK} = z - \frac {1} {2} (x+y)?$$ – Dbchatto67 Apr 22 at 11:16
• 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. – Dbchatto67 Apr 22 at 11:24

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.