Problem — Varignon's parallelogram and its classifications

Text

Let $EFGH$ be an arbitrary quadrilateral and let $P$, $Q$, $R$, $S$ be the midpoints of the sides $EF$, $FG$, $GH$, $HE$ respectively.

• (a) Prove that $PQRS$ is a parallelogram.
• (b) Determine under which conditions on the diagonals $EG$ and $FH$ of the original quadrilateral the parallelogram $PQRS$ is a rectangle, a rhombus, a square.

(original figure: quadrilateral $EFGH$ with the midpoints of the sides and the parallelogram $PQRS$)

Solution

(a) Draw the diagonal $EG$. In triangle $EFG$ the points $P$ and $Q$ are the midpoints of the sides $EF$ and $FG$: by the midsegment theorem $PQ \parallel EG$ and $PQ \cong \tfrac12 EG$. In triangle $GHE$ the points $R$ and $S$ are the midpoints of the sides $GH$ and $HE$: hence $SR \parallel EG$ and $SR \cong \tfrac12 EG$. It follows that $PQ \parallel SR$ and $PQ \cong SR$: having one pair of opposite sides parallel and congruent, $PQRS$ is a parallelogram. $\;\blacksquare$ Drawing the other diagonal $FH$ one obtains similarly $QR \parallel FH$, $PS \parallel FH$ and $QR \cong PS \cong \tfrac12 FH$.

(b) The consecutive sides $PQ$ and $QR$ of the parallelogram are parallel to the diagonals $EG$ and $FH$ respectively, and measure $\tfrac12 EG$ and $\tfrac12 FH$. Therefore:

• $PQRS$ is a rectangle $\iff$ $PQ \perp QR$ $\iff$ $EG \perp FH$: the diagonals of the quadrilateral are perpendicular.
• $PQRS$ is a rhombus $\iff$ $PQ \cong QR$ $\iff$ $\tfrac12 EG \cong \tfrac12 FH$ $\iff$ $EG \cong FH$: the diagonals are congruent.
• $PQRS$ is a square $\iff$ both conditions hold: the diagonals of the quadrilateral are perpendicular and congruent. $\boxed{PQRS\ \text{parallelogram};\quad \text{rectangle}\iff EG\perp FH,\ \ \text{rhombus}\iff EG\cong FH,\ \ \text{square}\iff\text{both}}$