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

In the figure, $ABCD$ is a kite in which $AB=AD$ and $BC=CD$

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

To prove: $PQRS is a rectangle.

Construction: Join $AC$ and $BD$.

Proof: In $\Delta ABD,$

$P$ and $S$ are mid points of $AB$ and $AD$

By Mid Point Theorem,

$\therefore PS||BD$ and $PS=\frac{1}{2}BD$ ....(i)

Similarly in $\Delta BCD$, using Mid Point Theorem,

$Q$ and $R$ the mid points of $BCand$CD$

$\therefore QR||BDandQR=\frac{1}{2}BD$ ....(ii)

Form (i) and (ii),

$PS||QR$ and $PS=QR$

Therefore, $PQRS$ is a parallelogram [As one pair of opposite is equal and parallel.]

Now, In $△ADC$, $S$ and $R$ are mid points of of sides $AD$ and $CD$.

Thus, using mid point theorem,

$SR||AC$ ....(iii)

From (ii) and( iii), quadrilateral $ESFO$ is parallelogram. [As opposite sides are parallel].

Now, we know that kite is a special type of parallelogram and thus its diagonal bisect each other at ${90}^{\circ }$.

Therefore, in parallelogram $ESFO$,

$\angle ESF=\angle FOE={90}^{\circ }$ [Opposite angles are equal in a parallelogram.]

As one angle of parallelogram $PQRS$ is ${90}^{\circ }$, hence,

$PQRS$ is a rectangle.