Is $EFGH$ a parallelogram in quadrilateral $ABCD$?

[Ackbach]

Here is this week's POTW:

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Given a quadrilateral $ABCD$ with respective midpoints $EFGH$, show that the quadrilateral $EFGH$ is a parallelogram.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

Congratulations to Rido12 for his correct solution, which follows:

Spoiler

Let ABCD be a quadrilateral with respective midpoints $EFGH$. Then $\vec{EF}=\vec{EB}+\vec{BF}=\frac{1}{2}\vec{AB}+\frac{1}{2}\vec{BC}=\frac{1}{2}(\vec{AB}+\vec{BC})=\frac{1}{2}\vec{AC}$. Also, $\vec{HG}=\vec{HD}+\vec{DG}=\frac{1}{2}(\vec{AD}+\vec{DC})=\frac{1}{2}\vec{AC}$. Therefore $\vec{EF}$ and $\vec{HG}$ are of equal length and parallel.

Similarly, $\vec{FG}=\vec{FC}+\vec{CG}=\frac{1}{2}(\vec{BC}+\vec{CD})=\frac{1}{2}\vec{BD}$. And, $\vec{EH}=\vec{EA}+\vec{AH}=\frac{1}{2}(\vec{BA}+\vec{AD})=\frac{1}{2}\vec{BD}$.
Hence, $\vec{FG}$ and $\vec{EH}$ are parallel and equal, and $EFGH$ forms a parallelogram.