Vectors in a rectangle Let $ABCD$ be a rectangle , $K$ a point on $AC$ such that $BK$ is perpendicular to $AC$, and $M$, $N$ are the midpoints of $AK, CD$.  Use vectors to show that $MB$ is orthogonal to $MN$.
I know that if $BK$ is perpendicular to $AC$ that the dot product is $0$.  I also know that I can write the magnitude of vectors in terms of other vectors.  So what I keep getting to is:
$\vec{MB}=\vec{MC}+\vec{CB}=\vec{MA}+\vec{AB}=\vec{MA}+2\vec{NC}$
$\vec{MN}=\vec{MC}-\vec{NC}$
$\vec{BK}=-\vec{AB}+2\vec{AM}$
$\vec{AC}=\vec{BC}-\vec{AB}$
I can not find a connection between these vectors to show that the dot product of $\vec{MB}$ and $\vec{MN}$ equals $0$.  
 A: Let $\overrightarrow{AB}=\bar u$ and $\overrightarrow{AD}=\bar v$. Then $\overrightarrow{AC}=\bar u+\bar v$.
$\overrightarrow{AK}=k\overrightarrow{AC}$ for some $k\in\mathbb{R}$.
Note that $\overrightarrow{AC}\cdot\overrightarrow{BK}=0$.
\begin{align}
\overrightarrow{AC}\cdot(\overrightarrow{AK}-\overrightarrow{AB})&=0\\
(\bar u+\bar v)\cdot(k\bar u+k\bar v-\bar u)&=0\\
(k-1)|\bar u|^2+k|\bar v|^2&=0\\
k&=\frac{|\bar u|^2}{|\bar u|^2+|\bar v|^2}
\end{align}
We have $\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AK}$ and hence
\begin{align}
\overrightarrow{MB}&=\overrightarrow{AB}-\overrightarrow{AM}\\
&=\overrightarrow{AB}-\frac{1}{2}\overrightarrow{AK}\\
&=\bar u-\frac{k}{2}(\bar u+\bar v)\\
&=\left(1-\frac{k}{2}\right)\bar u-\frac{k}{2}\bar v
\end{align}
We have
\begin{align}
\overrightarrow{MN}&=\overrightarrow{AN}-\overrightarrow{AM}\\
&=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}-\frac{1}{2}\overrightarrow{AK}\\
&=\bar v+\frac{1}{2}\bar u-\frac{k}{2}(\bar u+\bar v)\\
&=\left(\frac{1}{2}-\frac{k}{2}\right)\bar u+\left(1-\frac{k}{2}\right)\bar v
\end{align}
Therefore,
\begin{align}
\overrightarrow{MB}\cdot\overrightarrow{MN}&=\left(1-\frac{k}{2}\right)\left(\frac{1}{2}-\frac{k}{2}\right)|\bar u|^2-\frac{k}{2}\left(1-\frac{k}{2}\right)|\bar v|^2\\
&=\left(\frac{1}{2}-\frac{3k}{4}+\frac{k^2}{4}\right)|\bar u|^2+\left(-\frac{k}{2}+\frac{k^2}{4}\right)|\bar v|^2\\
&=\frac{1}{2}|\bar u|^2-\frac{k}{4}(3|\bar u|^2+2|\bar v|^2)+\frac{k^2}{4}(|\bar u|^2+|\bar v|^2)\\
&=\frac{1}{2}|\bar u|^2-\frac{|\bar u|^2(3|\bar u|^2+2|\bar v|^2)}{4(|\bar u|^2+|\bar v|^2)}+\frac{|\bar u|^4}{4(|\bar u|^2+|\bar v|^2)}\\
&=\frac{|\bar u|^2(2|\bar u|^2+2|\bar v|^2-3|\bar u|^2-2|\bar v|^2+|\bar u|^2)}{4(|\bar u|^2+|\bar v|^2)}\\
&=0
\end{align}
A: Let $AB=a$, $AD=b$, $\vec{AB}=\vec{u}$ and $\vec{AD}=\vec{v}$.
Thus,
$$\vec{BM}=-\vec{u}+\frac{AK}{2AC}\cdot\vec{AC}=$$
$$=-\vec{u}+\frac{a^2}{2(a^2+b^2)}(\vec{u}+\vec{v})=-\frac{a^2+2b^2}{2(a^2+b^2)}\vec{u}+\frac{a^2}{2(a^2+b^2)}\vec{v}$$ and
$$\vec{MN}=-\frac{a^2}{2(a^2+b^2)}(\vec{u}+\vec{v})+v+\frac{1}{2}\vec{u}=\frac{b^2}{2(a^2+b^2)}\vec{u}+\frac{a^2+2b^2}{2(a^2+b^2)}\vec{v}.$$
Id est, $$\vec{BM}\cdot\vec{MN}=\left(-\frac{a^2+2b^2}{2(a^2+b^2)}\vec{u}+\frac{a^2}{2(a^2+b^2)}\vec{v}\right)\left(\frac{b^2}{2(a^2+b^2)}\vec{u}+\frac{a^2+2b^2}{2(a^2+b^2)}\vec{v}\right)=$$
$$=-\frac{(a^2+2b^2)a^2b^2}{4(a^2+b^2)^2}+\frac{(a^2+2b^2)a^2b^2}{4(a^2+b^2)^2}=0$$
and we are done!
There is a nice proof without vectors. 
