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.

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


*

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

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

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


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

