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 $0$.  

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. 
