Question

ABCD is a rectangle and $P,Q,R$ and $S$ are mid -points of the $AB,BC,CD$ and $DA$ respectively. Show that the quadrilateral PQRS is a rhombus.

Answer 1

Let us join $AC$ and $BD$

In $△ABC$

$P$ and $Q$ are the midpoints of AB and BC respectively

$\therefore PQ\parallel ACandPQ=\frac{1}{2}AC⟶\left(1\right)$ (from midpoint theorem )

similarly in $△ADC$

$SR\parallel ACandSR=\frac{1}{2}AC⟶\left(2\right)$

clearly, $PQ\parallel SRandPQ=SR$

since in quadrilateral $PQRS$, one pair of opposite side is equal and parallel to each other then it is a parallelogram.

$PQ\parallel QRandPS=QR⟶\left(3\right)$ (opposite side of parrallelogram)

In $△BCD$, $Q$ and $R$ are the midpoints of side $BC$ and $CD$ respectively

$\therefore QR\parallel BDandQR=\frac{1}{2}BD⟶\left(3\right)$ (by midpoint theorem)

since we know that the diagonals of a rectangle are equal

$\therefore AC=BD⟶\left(5\right)$

By using equation $\left(1\right),\left(2\right),\left(3\right),\left(4\right),\left(5\right)$ we obtain

$PQ=QR=SR=RS$

$\therefore PQRS$ is a rhombus