Question

ABCD is a rhombus and P,Q,R,S are joined with the midpoints of AB,BC,CD,DA respectively . Show that the quadrilateral PQRS is a rectangle

Answer 1

Given- ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

To Prove-PQRS is a rectangle:

Construction,
AC and BD are joined.

Proof,
In ΔDRS and ΔBPQ,
DS = BQ (Halves of the opposite sides of the rhombus)
∠SDR = ∠QBP (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
Thus, ΔDRS ≅ ΔBPQ by SAS congruence condition.
RS = PQ by CPCT --- (i)
In ΔQCR and ΔSAP,
RC = PA (Halves of the opposite sides of the rhombus)
∠RCQ = ∠PAS (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
Thus, ΔCQR ≅ ΔAPS by SAS congruence condition.
RQ = SP by CPCT --- (ii)
Now,
In ΔCDB,
R and Q are the mid points of CD and BC respectively.
⇒ QR || BD
also,
P and S are the mid points of AD and AB respectively.
⇒ PS || BD
⇒ QR || PS
Thus, PQRS is a parallelogram.
To prove at least one angle is ${90}^{\circ }$:
Since, $AB=BC$ as sides of rhombus are equal.
$\Rightarrow \frac{AB}{2}=\frac{BC}{2}$
$\Rightarrow PB=BQ$
In $\Delta BPQ$
$\angle 1=\angle 2$ as $PB=BQ$
Since, $\Delta APS\cong \Delta CQR$
So, $\angle APS=\angle CQR=\angle 3$
As AB is straight line,
So $\angle 1+\angle P+\angle 3={180}^{\circ }$.......(iii) and
BC is also a straight line
$\Rightarrow \angle 2+\angle Q+\angle 3={180}^{\circ }$ .....(iv)
From equation (iii) and (iv), we get
$\Rightarrow \angle 2+\angle Q+\angle 3=\angle 1+\angle P+\angle 3$
$\therefore \angle P=\angle Q$
As $PS\parallel QR$ and $PQ$ is a transversal.
So, $\angle P+\angle Q={180}^{\circ }$
$\Rightarrow 2\angle P=180$ or $2\angle Q=180$
$\therefore \angle P={90}^{\circ }$ or $\angle Q={90}^{\circ }$

Therefore, ∠PQR = 90°
Now, In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
and also ∠Q = 90°
Thus, PQRS is a rectangle.