Problem — The midpoints of a rectangle's sides form a rhombus

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Let $KLMN$ be a rectangle and let $P$, $Q$, $R$, $S$ be the midpoints of the sides $KL$, $LM$, $MN$, $NK$ respectively. Prove that the quadrilateral $PQRS$ is a rhombus.

(original figure: rectangle $KLMN$ with the midpoints of the sides and the quadrilateral $PQRS$)

Solution

Let $2a$ denote the length of the sides $KL$ and $MN$, and $2b$ that of the sides $LM$ and $NK$ (in a rectangle opposite sides are congruent). Since $P,Q,R,S$ are midpoints: $KP\cong PL\cong MR\cong RN\cong a,\qquad LQ\cong QM\cong NS\cong SK\cong b.$ Consider the four triangles formed at the vertices of the rectangle: $PLQ$, $QMR$, $RNS$, $SKP$. Each has a right included angle (the angles at $L$, $M$, $N$, $K$ are right, being angles of the rectangle) and its two legs congruent to $a$ and $b$:

• $PLQ$: legs $PL\cong a$, $LQ\cong b$;
• $QMR$: legs $MR\cong a$, $QM\cong b$;
• $RNS$: legs $RN\cong a$, $NS\cong b$;
• $SKP$: legs $KP\cong a$, $SK\cong b$.

By the SAS congruence criterion (two sides and the included angle) the four triangles are congruent to one another. Consequently their hypotenuses are congruent: $PQ\cong QR\cong RS\cong SP.$ The quadrilateral $PQRS$ has its four sides congruent, hence it is a rhombus. $\;\blacksquare$