Prove quadrilateral ABCD is a parallelogram

Question

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In the image, EFGH is a parallelogram, and BE=HC=GD=AF.
Can I prove that ABCD is also a parallelogram?

Answer 1

I think the question should be reversed.

You are given $ABCD$ parallelogram and points $E,F,G,H$ are distributed on the sides with equal distance.

We can use the vectorial characterisation of a parallelogram $\vec{AD}=\vec{BC}$.

Then with Chasles additive relations:

$\vec{FG}=\underbrace{\vec{FA}}_{=\vec{CH}}+\underbrace{\vec{AD}}_{=\vec{BC}}+\underbrace{\vec{DG}}_{=\vec{EB}}=\vec{EB}+\vec{BC}+\vec{CH}=\vec{EH}$

So $EHGF$ is a parallelogram too.

• We can say $\vec{FA}=\vec{CH}$ because $(AD)//(BC)$ and $F\in(AB)$ and $H\in(BC)$ and $AF=CH$ in distance.

Similar reason for $\vec{DG}=\vec{EB}$.