How to prove (vectorially) that the perpendicular bisectors of the sides of a triangle are concurrent?
My Attempt:
Let $\triangle OAB$ be our triangle, let us take the positive $x$-axis along the direction from point $O$ to point $A$, let us take the length of the side $\overline{OA}$ to be our unit. Then we have $$ \vec{OA} = \hat{i} \ \ \ \mbox{ and } \ \ \ \vec{OB} = \mu \hat{i} + \nu \hat{j}, \tag{0} $$ where $\mu$ and $\nu$ are some (real) constants such that $\nu \neq 0$.
Then we have $$ \begin{align} \vec{AB} &= \vec{AO} + \vec{OB} \\ &= - \vec{OA} + \vec{OB} \\ &= - \hat{i} + \left( \mu \hat{i} + \nu \hat{j} \right) \\ &= (\mu - 1 ) \hat{i} + \nu \hat{j}, \end{align} $$ that is, we have $$ \vec{AB} = (\mu - 1 ) \hat{i} + \nu \hat{j}. \tag{1} $$
Let $C$, $D$, and $E$, respectively, be the midpoints of the sides $\overline{OA}$, $\overline{OB}$, and $\overline{AB}$.
Then we have $$ \begin{align} \vec{OC} &= \frac12 \vec{OA} = \frac12 \hat{i}, \qquad \mbox{[ using (0) above ]} \\ \vec{OD} &= \frac12 \vec{OB} = \frac12 \left( \mu \hat{i} + \nu \hat{j} \right), \qquad \mbox{[ using (0) above ]} \\ \vec{OE} &= \vec{OA} + \vec{AE} = \hat{i} + \frac12 \vec{AB} \qquad \mbox{[ using (0) above ]} \\ &= \hat{i} + \frac12 \left( (\mu - 1 ) \hat{i} + \nu \hat{j} \right) \qquad \mbox{[ using (1) above ]} \\ &= \frac{ \mu + 1}{2} \hat{i} + \frac{\nu}{2} \hat{j}. \end{align} \tag{2} $$
Let $P$ be the point of intersection of the perpendicular bisectors of the sides $\overline{OA}$ and $\overline{OB}$, and let $Q$ be the point of intersection of the perpendicular bisectors of the sides $\overline{OA}$ and $\overline{AB}$.
We need to show that these points $P$ and $Q$ coincide. For this purpose we show that $$ \vec{OP} = \vec{OQ}. \tag{3} $$
As $\overline{CP} \perp \overline{OA}$ and as $\overline{OA}$ is parallel to our $x$-axis, so $\overline{CP}$ is parallel to the $y$-axis, which implies that $$ \vec{CP} = y \hat{j}, $$ where $y$ is some (real) constant to be determined, and hence $$ \begin{align} \vec{OP} &= \vec{OC} + \vec{CP} \\ &= \frac12 \hat{i} + y \hat{j} \qquad \mbox{[ using (2) above ]}, \end{align} $$ that is, we have $$ \vec{OP} = \frac12 \hat{i} + y \hat{j}. \tag{4} $$
Then we have $$ \begin{align} \vec{DP} &= \vec{DO} + \vec{OP} \\ &= -\vec{OD} + \vec{OP} \\ &= - \frac12 \left( \mu \hat{i} + \nu \hat{j} \right) + \left( \frac12 \hat{i} + y \hat{j} \right) \qquad \mbox{[ using (2) and (4) above ]} \\ &= \frac{ 1 - \mu }{2} \hat{i} + \frac{2y - \nu }{2 } \hat{j}. \end{align} \tag{5} $$
Now since $\overline{DP} \perp \overline{OB}$, we have $$ \vec{DP} \cdot \vec{OB} = 0, $$ that is, $$ \left( \frac{ 1 - \mu }{2} \hat{i} + \frac{2y - \nu }{2 } \hat{j} \right) \cdot \left( \mu \hat{i} + \nu \hat{j} \right) = 0, $$ or in other words, $$ \frac{ (1-\mu)\mu + (2y-\nu)\nu }{2} = 0, $$ which implies $$ (1-\mu)\mu + (2y-\nu)\nu = 0, $$ and hence $$ y = \frac12 \left( \nu + \frac{ (\mu - 1)\mu }{\nu} \right) = \frac{ (\mu - 1)\mu + \nu^2}{2 \nu} . \tag{6} $$
Therefore by (4) above and (6) we have $$ \vec{OP} = \frac12 \hat{i} + y \hat{j} = \frac12 \hat{i} + \frac{ (\mu - 1)\mu + \nu^2}{2\nu} \hat{j}. \tag{7} $$
Now we show that $\overline{PE} \perp \overline{AB}$.
We see that $$ \begin{align} \vec{PE} &= \vec{PO} + \vec{OE} \\ &= -\vec{OP} + \vec{OE} \\ &= - \left( \frac12 \hat{i} + \frac{ (\mu - 1)\mu + \nu^2}{2\nu} \hat{j} \right) + \left( \frac{ \mu + 1}{2} \hat{i} + \frac{\nu}{2} \hat{j} \right) \qquad \mbox{[ using (7) and (2) above ]} \\ &= \frac{\mu}{2}\hat{i} + \frac{ (1-\mu)\mu }{2\nu} \hat{j}, \end{align} $$ that is, $$ \vec{PE} = \frac{\mu}{2}\hat{i} + \frac{ (1-\mu)\mu }{2\nu} \hat{j}. \tag{8} $$ Therefore we have $$ \begin{align} \vec{PE} \cdot \vec{AB} &= \left( \frac{\mu}{2}\hat{i} + \frac{ (1-\mu)\mu }{2\nu} \hat{j} \right) \cdot \left( (\mu - 1 ) \hat{i} + \nu \hat{j} \right) \qquad \mbox{[ using (8) and (1) above ]} \\ &= \frac{ \mu ( \mu - 1 ) }{2} + \frac{(1-\mu)\mu }{2\nu} \nu \\ &= \frac{ \mu ( \mu - 1 ) }{2} + \frac{(1-\mu)\mu }{2} \\ &= 0, \end{align} $$ thus showing that $\overline{PE} \perp \overline{AB}$. Therefore the perpendicular bisector of $\overline{AB}$ also passes through the point $P$, which is the point of intersection of the perpendicular bisectors of the sides $\overline{OA}$ and $\overline{OB}$.
Am I right?
Alternatively, using the same reasoning as in (4) above, we have $$ \vec{OQ} = \frac12 \hat{i} + z \hat{j}, \tag{10} $$ where $z$ is some (real) constant to be determined.
Then $$ \begin{align} \vec{QE} &= \vec{QO} + \vec{OE} \\ &= -\vec{OQ} + \vec{OE} \\ &= - \left( \frac12 \hat{i} + z \hat{j} \right) + \left( \frac{ \mu + 1}{2} \hat{i} + \frac{\nu}{2} \hat{j} \right) \qquad \mbox{[ using (10) and (2) above ]} \\ &= \frac{\mu}{2} \hat{i} +\frac{\nu - 2z}{2} \hat{j}, \end{align} $$ that is, $$ \vec{QE} = \frac{\mu}{2} \hat{i} +\frac{\nu - 2z}{2} \hat{j}. \tag{11} $$
Now since $\overline{QE} \perp \overline{AB}$, therefore we have $$ \vec{QE} \cdot \vec{AB} = 0, $$ that is, $$ \left( \frac{\mu}{2} \hat{i} +\frac{\nu - 2z}{2} \hat{j} \right) \cdot \left( (\mu - 1 ) \hat{i} + \nu \hat{j} \right) = 0, $$ [Refer to (11) and (1) above.] or in other words, $$ \frac{\mu(\mu-1)}{2} + \frac{(\nu - 2z)\nu}{2} = 0, $$ which implies $$ \mu(\mu-1) + (\nu - 2z)\nu = 0, $$ and hence $$ z = \frac{1}{2} \left( \nu + \frac{\mu (\mu - 1) }{\nu} \right) = \frac{\mu(\mu-1) + \nu^2 }{2\nu}, $$ and then from (10) we have $$ \vec{OQ} = \frac12 \hat{i} + \frac{\mu(\mu-1) + \nu^2 }{2\nu} \hat{j}. \tag{12} $$
Thus from (7) and (12) above, we can conclude that (3) holds and our desired conclusion follows.
Am I right?
Is my proof correct and clear enough in each and every detail? If so, is my presentation easy to understand? Or, are there any issues of accuracy, detail, or clarity?
Let the triangle be $ABC$ with $P, Q$ and $R$ the midpoints of the sides opposite the respective vertices.
Let $\overrightarrow{AR}=\underline{a}=\overrightarrow{RB}$ and $\overrightarrow{AQ}=\underline{b}=\overrightarrow{QC}$
Let $O$ be the intersection of the perpendicular bisectors of $AB$ and $AC$, so that $\overrightarrow{RO}=\underline{u}$ and $\overrightarrow{QO}=\underline{v}$.
It then follows that: $$\underline{a}\cdot\underline{u}=0=\underline{b}\cdot\underline{v}$$ And that $$\overrightarrow{AO}=\underline{a}+\underline{u}=\underline{b}+\underline{v}$$
It requires to be shown that $$\overrightarrow{OP}\perp\overrightarrow{BC}$$
Now $\overrightarrow{BP}=\underline{b}-\underline{a}$, so $\overrightarrow{OP}=-\underline{u}+\underline{b}$
Then we have $$\overrightarrow{OP}\cdot\overrightarrow{BC}=(-\underline{u}+\underline{b})\cdot(2\underline{b}-2\underline{a})$$ $$=2\underline{b}\cdot(-\underline{u}+\underline{b}- \underline{a})$$ $$=2\underline{b}\cdot(-\underline{v})=0$$
Hence proved.