Question

$ABCD$ is a kite having $AB=AD$ and $BC=CD$. Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.

Answer 1

Given:A kite $ABCD$ having $AB=AD$ and $BC=CD$

$P,Q,R,S$ are the midpoint of sides $AB,BC,CD$, and $DA$ respectively.

$PQ,QR,RS$ and spare joined

To prove: $PQRS$ is a rectangle

Proof:In $△ABC,P$ and $Q$ are the midpoints of $AB$ and $BC$ respectively.

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

In $△ADC,R$ and $S$ are the midpoint of $CD$ and $AD$ respectively.

$\therefore RS\parallel AC$ and $RS=\frac{1}{2}AC$ ....$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we have

$PQ\parallel RS$ and $PQ=RS$

Thus, in quadrilateral $PQRS$, a pair of opposite sides are equal and parallel. So $PQRS$ is a parallelogram.

Now, we shall prove that one angle of parallelogram $PQRS$ it is a right angle.

Since $AB=AD$

$\Rightarrow \frac{1}{2}AB=\frac{1}{2}AD$

$\Rightarrow AP=AS$ ........$\left(3\right)$

$\Rightarrow \angle 1=\angle 2$ ......$\left(4\right)$

Now,in $△PBO$ and $△SDR$ we have

$PB=SD$ since $AD=AB\Rightarrow \frac{1}{2}AD=\frac{1}{2}AAB$

$BQ=DR$

$\therefore PB=SD$

And $PQ=SR$ since $PQRS$ is a parallelogram.

So by SSS criterion of congruence, we have

$△PBQ\cong △SOR$

$\Rightarrow \angle 3=\angle 4$ by CPCT

Now, $\angle 3+\angle SPQ+\angle 2={180}^{\circ }$

And $\angle 1+\angle PSR+\angle 4={180}^{\circ }$

$\therefore \angle 3+\angle SPQ+\angle 2=\angle 1+\angle PSR+\angle 4$

$\Rightarrow \angle SPQ=\angle PSR$ since $\angle 1=\angle 2$ and $\angle 3=\angle 4$

Now, transversal $PS$ cuts parallel lines $SR$ and $PQ$ at $S$ and $P$ respectively.

$\therefore \angle SPQ+\angle PSR={180}^{\circ }$

$\Rightarrow 2\angle SPQ={180}^{\circ }$

$\Rightarrow \angle SPQ=\frac{{180}^{\circ }}{2}={90}^{\circ }$

Thus, $PQRS$ is a parallelogram such that $\angle SPQ={90}^{\circ }$

Hence, $PQRS$ is a parallelogram.