Question

$P,Q,R,\mathrm{and}S$ are respectively the mid-points of the sides $AB,BC,CD,\mathrm{and}DA$ of a quadrilateral $ABCD$ such that $AC\perp BD$. Prove that $PQRS$ is a rectangle.

Answer 1

Step 1. Prove that opposite sides are equal and parallel.

It is given that $P,Q,R,\mathrm{and}S$ are respectively the mid-points of the sides $AB,BC,CD,\mathrm{and}DA$ of a quadrilateral $ABCD$.

Apply the mid-point theorem in $∆ABD$ and in $∆BDC$.

In $∆ABD$, $PS=\frac{1}{2}BD$ and $PS\parallel BD$.

In $∆BDC$, $QR=\frac{1}{2}BD$ and $QR\parallel BD$.

$\Rightarrow PS=QR=\frac{1}{2}BD\cdots \left(1\right)$ and $PS\parallel QR$.

Again, apply the mid-point theorem in $∆ABC$ and in $∆ADC$.

In $∆ABC$, $PQ=\frac{1}{2}AC$ and $PQ\parallel AC$.

In $∆ADC$, $SR=\frac{1}{2}AC$ and $SR\parallel AC$.

$\Rightarrow PQ=SR=\frac{1}{2}AC$ and $PS\parallel QR$.

Hence, the opposite sides are equal and parallel.

Step 2. Prove that the parallelogram forms a right angle.

Since it is given that $AC\perp BD$, $\angle COD=\angle AOD=\angle AOB=\angle COB={90}^{°}$.

Assume the quadrilateral $TOUS$.

Thus, $OT\parallel SU\mathrm{and}ST\parallel OU$.

$\begin{array}{rcl}\angle TOU& =& {90}^{°}\left[\therefore \angle AOD={90}^{°}\right]\end{array}$

So, $\begin{array}{rcl}\angle TOU& =& \angle TSU={90}^{°}\end{array}$ (Opposite angles of the parallelogram)

$\begin{array}{rcl}& \Rightarrow & \angle PSR={90}^{°}\end{array}$

Thus, the parallelogram forms a right angle.

Hence, $PQRS$ is a rectangle.