Prove vectorially that the perpendicular bisectors of the sides of a triangle are concurrent 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?
 A: 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.
