Question

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

Answer 1

Step 1. Prove that $PQRS$ has equal four sides.

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

It is also given that $AC=BD$ and $AC\perp BD$.

Apply the mid-point theorem in $∆ABC$ and $∆ADC$.

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

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

$\Rightarrow PQ\parallel SR$ and $PQ=SR=\frac{1}{2}AC\cdots \left(1\right)$.

Again, apply the mid-point theorem in $∆ABD$ and $∆BDC$.

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

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

$\Rightarrow PS\parallel QR$ and $PS=QR=\frac{1}{2}BD$.

Since $AC=BD$, $PS=QR=\frac{1}{2}AC\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, $PQ=SR=PS=QR$.

Hence, all four sides of $PQRS$ are equal.

Step 2. Prove that $PQRS$ is a square.

Assume the quadrilateral $TSUO$.

$OT\parallel SU,ST\parallel OU$

Since $AC\perp BD$, $\angle TOU={90}^{°}$.

$\angle TOU=\angle TSU={90}^{°}$ (Opposite angles of the parallelogram)

$\Rightarrow \angle PSR={90}^{°}$

Similarly all the angle of $PQRS$ are right angles.

Hence, $PQRS$ is a square.