Prove that $ABCD$ is a rectangle. I am looking for a synthetic proof Let $ABCD$ a  parallelogram  and $ BE \perp AC, E \in (AC)$, $M$  is  the middle of $[AE]$  and  $N$  is  the  middle of $[CD]$.  If $ BM \perp NM$ then $ABCD$  is   a  rectangle. I   try  to  show  that  $MNCB$  is  inscriptible. I am looking for a synthetic proof.
 A: Here is a synthetic proof: 
Let $H$ be the midpoint of segment $BE$. Then $MH$ is a midsegment of right-angled triangle $ABE$ parallel to the hypotenuse $AB$ and $MH = \frac{1}{2}AB$. Therefore, 
$$MH = \frac{1}{2}AB = \frac{1}{2}CD = NC$$ and since $MH$ is parallel to $AB$, which in its own turn is parallel to $CD$, one concludes that $MH$ is parallel and equal in length to $NC$. Therefore $MHCN$ is a parallelogram which means that $MN$ is parallel to  $CH$. However, $MN$ is orthogonal to $MB$ hence $CH$ is also orthogonal to $BM$. Thus, lines $CH$ and $BE$ are two out of the three altitudes in triangle $BCM$ intersecting at $H$. Thus, $H$ is the orthocenter of triangle $BCM$. Consequently, line $MH$ is the third altitude and is therefore orthogonal to $BC$. However, $MH$ is parallel to both $AB$ and $CD$ which means that both $AB$ and $CD$ are orthogonal to $BC$, i.e. $ABCD$ is a rectangle.  

A: $$\overrightarrow{BM}\cdot \overrightarrow{MN} = 0$$
$$ \Leftrightarrow (\overrightarrow{BA}\cdot \overrightarrow{BE})\cdot(\overrightarrow{AD} \cdot \overrightarrow{EC}) = 0$$
$$ \Leftrightarrow \color\red{\overrightarrow{BA}\cdot \overrightarrow{AD}} + \color\RoyalBlue{\overrightarrow{BA} \cdot \overrightarrow{EC}} + \color\green{\overrightarrow{BE} \cdot \overrightarrow{AD}} + \overrightarrow{BE}\cdot \overrightarrow{EC}= 0$$


*

*$\color\red{\overrightarrow{BA} \cdot \overrightarrow{AD}  = \overrightarrow{BA} \cdot \overrightarrow{BC} = (\overrightarrow{BE} + \overrightarrow{EA})\cdot(\overrightarrow{BE}+\overrightarrow{EC}) = \overrightarrow{BE}^2 + \overrightarrow{EA}\cdot {EC}}$

*$\color\RoyalBlue{\overrightarrow{BA} \cdot \overrightarrow{EC}  = (\overrightarrow{BE}+\overrightarrow{EA})\cdot \overrightarrow{EC} = \overrightarrow{EA} \cdot \overrightarrow{EC}}$

*$\color\green{\overrightarrow{BE}\cdot \overrightarrow{AD} = \overrightarrow{BE} \cdot \overrightarrow{BC} = \overrightarrow{BE}\cdot (\overrightarrow{BE} + \overrightarrow{EC}) = \overrightarrow{BE}^2}$


$$\Longrightarrow 2(\overrightarrow{BE}^2 + \overrightarrow{EA}\cdot \overrightarrow{EC}) = 0$$
$$\Rightarrow BE^2 = EA\times EC$$
$$\Rightarrow \angle{ABC} = 90^\circ.$$

