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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.

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    It is not very polite to use three exclamation marks in the question. It's like shouting in a conversation. – 2017-01-29

2 Answers 2

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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.

enter image description here

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

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

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

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    You have $M$ and $N$ switched in your diagram. Does that affect the answer? – 2017-01-29
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    @SSepher Great answer. I am looking for a synthetic proof. – 2017-01-29
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    No, the solution is according to my figure. – 2017-01-29
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    Your first line should then be $\vec{BN}\cdot\vec{MN}=0$. – 2017-01-29