In the following figure, $ABCD$ is a parallelogram, and $O$ is any point. Parallelograms $OAEB, OBFC, OCGD, ODHA$ are completed. Prove that $EFGH$ is a parallelogram.
We can obtain a fairly trivial proof using affine geometry. As $OAEB, OBFC, OCGD, ODHA$ are parallelograms,
$$\vec{A} + \vec{B} = \vec{O} + \vec{E} \implies \vec{A} + \vec{B} – \vec{O} = \vec{E} \tag1$$
$$\vec{C} + \vec{B} – \vec{O} = \vec{F}\tag2$$
$$\vec{D} + \vec{C} – \vec{O} = \vec{G} \tag3$$
$$\vec{D} + \vec{A} – \vec{O} = \vec{H}\tag4$$
Adding $(1)$ to $(3)$ and $(2)$ to $(4)$, we get,
$$\vec{E} + \vec{G} = \vec{A} + \vec{B} + \vec{C} + \vec{D} – 2\vec{O} \tag5$$
$$\vec{F} + \vec{H} = \vec{A} + \vec{B} + \vec{C} + \vec{D} – 2\vec{O} \tag6$$
Clearly, $(5)$ and $(6)$ are equal, therefore,
$$\vec{E} + \vec{G} = \vec{F} + \vec{H}$$
Therefore, $EFGH$ is a parallelogram. Can somebody give an elementary proof using Euclidean geometry? Also, I noticed that in my proof, nowhere did I use the fact that $ABCD$ is a parallelogram, but constructing an example, it was quickly clear that the result stated does not generalize to all quadrilaterals. How come? Is my proof incorrect?
Best Answer
It is well-known (and easily provable) that the "mid-point polygon" of any quadrilateral is a parallelogram. (Opposite sides are parallel to a diagonal.) The Question's $EFGH$ is simply the dilation of the mid-point polygon with respect to point $O$, with scale factor $2$, and is therefore also a parallelogram.